Review a 250 Kg Sphere Is at the Origin and a 700 Kg Sphere Is at X=20cm
Momentum and Collisions Review
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Part East: Problem-Solving
57. A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/southward. It rebounds with a speed of 13.5 thousand/s. The contact time is 0.100 seconds. (a) determine the impulse with the wall, (b) make up one's mind the force of the wall on the brawl.
Answer: Answer: (a) -xvi.seven N s; (b) -167 N
Given: g = 0.530 kg; 5i = xviii.0 m/s; vf = xiii.5 grand/s; t = 0.100 s
Find: (a) Impulse, (b) Force
(a) Impulse = Momentum Change = thou•Delta v = thousand•(vf - fivei)= (0.530 kg)•( -13.5 chiliad/due south - 18.0 m/s)
Impulse = -16.7 kg•thousand/southward = -16.7 North•southward
where the "-" indicates that the impulse was opposite the original direction of motility.
(Note that a kg•m/s is equivalent to a N•s)
(b) The impulse is the product of force and time. So if impulse is known and time is known, force can be easily determined.
Impulse = F•t
F = Impulse/t = (-16.vii Northward south) / (0.100 south) = -167 N
where the "-" indicates that the impulse was opposite the original direction of motion.
58. A 4.0-kg object has a frontwards momentum of twenty. kg•thousand/s. A 60. N•due south impulse acts upon it in the direction of motion for five.0 seconds. A resistive force of 6.0 N and so impedes its motion for 8.0 seconds. Determine the final velocity of the object.
Respond: vf = 8.0 m/s
This question is all-time thought about conceptually using the principle that an objects momentum is changed when it encounters an impulse and the amount of change in momentum is equal to the impulse which it encounters.
Here an object starts with 20 units (kg•thousand/due south) of momentum. It then encounters an impulse of 60 units (North•s) in the management of motility. A 60-unit impulse will alter the momentum by 60 units, either increasing or decreasing it. If the impulse is in the direction of an object's motion, then it will increase the momentum. So now the object has 80 units (kg•m/s) of momentum. The object then encounters a resistive force of six.0 North for 8.0 s. This is equivalent to an impulse of 48 units (Due north•s). Since this impulse is "resistive" in nature, information technology will decrease the object's momentum past 48 units. The object now has 32 units of momentum. The question asks for the object's velocity after encountering these two impulses. Since momentum is the product of mass and velocity, the velocity can be easily determined.
5last = pterminal / m = (32 kg m/s) / (4.0 kg) = 8.0 m/s
59. A 3.0-kg object is moving forward with a speed of 6.0 thousand/s. The object and then encounters a force of two.5 N for 8.0 seconds in the direction of its motion. The object then collides head-on with a wall and heads in the opposite management with a speed of v.0 thou/due south. Determine the impulse delivered by the wall to the object.
Answer: 53 Due north•s
Similar the previous problem, this problem is best solved by thinking through it conceptually using the impulse-momentum change principle.
Here the object begins with a momentum of 18 units (kg•thousand/south). The object encounters a force of two.5 North for 8.0 seconds. This is equivalent to an impulse of xx units (Northward•s). Since this impulse acts in the management of motility, it changes the object's momentum from 18 units to 38 units. A concluding impulse is encountered when colliding with a wall. Upon rebounding, the object has a momentum of -15 units (kg•m/southward). The -15 is the product of mass (iii kg) and velocity (-five thousand/southward). The "-" sign is used since the object is now moving in the opposite direction as the original motion. The collision with the wall inverse the object's momentum from +38 units to -15 units. Thus, the collision must accept resulted in a 53-unit impulse since it altered the object's momentum past 53 units.
threescore. A 46-gram tennis ball is launched from a one.35-kg bootleg cannon. If the cannon recoils with a speed of 2.1 thou/s, determine the muzzle speed of the lawn tennis ball.
Answer: 62 m/s
Given: mbrawl = 46 g = 0.046 kg; one thousandcannon = one.35 kg; vcannon = -2.one m/s
Notice: vball = ???
The ball is in the cannon and both objects are initially at residual. The total system momentum is initially 0. After the explosion, the total system momentum must also exist 0. Thus, the cannon'due south astern momentum must be equal to the ball'due south forrard momentum.
(1.35 kg) • (-2.one thousand/s) = (0.046 kg) • fiveball
fivebrawl = (i.35 kg) • (2.one thou/s) / (0.046 kg) = 61.63 k/s = ~62 yard/s
61. A 2.0-kg box is attached by a cord to a 5.0-kg box. A compressed leap is placed between them. The ii boxes are initially at rest on a friction-free track. The string is cut and the bound applies an impulse to both boxes, setting them in motion. The 2.0-kg box is propelled backwards and moves 1.ii meters to the end of the track in 0.fifty seconds. Determine the time it takes the 5.0-kg box to motion 0.90 meters to the opposite end of the track.
Answer: 0.94 s
For the sake of the discussion, the 2-kg box will be referred to as Box ane and the 5-kg box will exist referred to as box 2.
Given: 1000box 1 = 2.0 kg; mbox 2 = five.0 kg; dbox 1 = i.2 g; tbox 1 = 0.50 south; dbox two = 0.90 m
Find: tbox 2 = ???
The two boxes are initially at rest. The full system momentum is initially 0. After the cutting of the string and the impulse of the bound, the total system momentum must also be 0. Thus, Box 1's backward momentum must be equal to the Box 2'due south forward momentum. The altitude and time for Box 1 must be used to determine its velocity.
Now the principle of momentum conservation can be used to determine Box 2's velocity.
(2 kg) • (two.4 m/southward) = (5 kg) • vbox 2
vbox 2 = (2 kg) • (2.4 m/s) / (5 kg) = 0.96 chiliad/s
The velocity of Box 2 can be used to make up one's mind the time it takes it to move a altitude of 0.90 meters.
Time = dbox 2 / fivebox 2 = (0.90 one thousand) / (0.96 thou/southward) = 0.9375 southward = ~0.94 s
62. 2 children are playing with a large snowball while on water ice skates on a frozen swimming. The 33-kg kid tosses the 5.0-kg snowball, imparting a horizontal speed of 5.0 g/s to it. The 33-kg child is four.0 meters from a 28-kg child and 8.0 meters from the edge of the swimming (located behind him). Assuming negligible friction, how much fourth dimension elapses between when the 28-kg kid gets hit by the snowball and when the 33-kg child reaches the edge of the swimming?
Answer: nine.8 s
For the sake of the give-and-take, nosotros volition refer to the 33-kg child as the "thrower" and the 28-kg child as the "catcher."
In this scenario, the thrower tosses a snowball forward towards the catcher. This throwing action involves an impulse imparted to the snowball. And due to action-reaction, in that location is an identical impulse imparted to the thrower which causes the thrower to be set in motility in the opposite management. The impulse is equal to the momentum change. And since the mass and the velocity modify of the snowball are known, the momentum change of the snowball can exist adamant.
yard • (Delta five)snowball = (5.0 kg) • (+ 5.0 m/due south - 0 m/s) = 25.0 kg•m/s
This 25-unit momentum modify of the snowball is equal to the thrower'south momentum change. Thus the thrower is moving backwards towards the edge of the swimming with a momentum of -25.0 kg•m/south. Since momentum is related to velocity, the post-impulse velocity tin can exist determined.
5thrower = pthrower / mthrower
5thrower = (-25.0 kg•m/s) / (33 kg)
vthrower = -0.7576 m/southward
The thrower began viii.0 meters from the border of the pond. Once the ball has been thrown, the thrower is moving backwards towards the border of the swimming with a speed of 0.7576 yard/s. Assuming negligible friction on the icy pond, the speed can exist used to determine the time that elapses between when the brawl is thrown and when the thrower reaches the pond's border.
t = d / five
tfor thrower to reach pond's edge = (eight.0 meters) / (0.7576 1000/s)
tfor thrower to attain pond'due south edge = x.56 seconds
One time the ball is thrown, the thrower starts moving backwards towards the pond'due south edge. Meanwhile, the ball is moving forrad towards the catcher. The fourth dimension for the ball to move from the thrower's original position to the catcher is dependent upon the ball'southward speed and the original distance betwixt the thrower and the catcher.
t = d / v
tfor ball to move from thrower to catcher = (four.0 meters) / (five.0 m/s)
tfor brawl to move from thrower to catcher = 0.800 seconds
The instant the ball is thrown, two motions occur - the ball moves forward towards the catcher and the thrower moves backwards towards the pond'due south edge. The brawl reaches the catcher in 0.800 seconds. And 9.76 seconds (~9.eight s) later (10.56 s - 0.80 s), the thrower reaches the pond's edge.
63. A 2.8-kg physics cart is moving forward with a speed of 45 cm/southward. A 1.9-kg brick is dropped from balance and lands on the cart. The cart and brick move together across the horizontal surface. Presume an isolated organization.
a. Decide the post-collision speed of the cart and the brick.b. Determine the momentum alter of the cart.
c. Decide the momentum modify of the brick.
d. Determine the internet impulse upon the cart.
e. Determine the net impulse upon the system of cart and brick.
Answers: (a) v = 27 cm/south
(b) Delta pcart = -51 kg • cm/s
(c) Delta pbrick = +51 kg • cm/s
(d) Impulse on cart = -51 kg • cm/s
(east) Impulse on brick = +51 kg • cm/s
Before the collision, just the moving cart has momentum. The total momentum of the system is simply the mass of the cart multiplied by the velocity of the cart.
ptotal-before = (2.viii kg) • (45 cm/s)
pfull-before = 126 kg • cm/s
The collision is perfectly inelastic; the ii objects stick together and move every bit a single unit. After the collision, the total momentum of the system is the sum of the individual momentum values.
pfull-afterward = mcart-after • vcart-after + mbrick • fivebrick-after
ptotal-afterward = one thousandcart-afterwards • v + chiliadbrick • v
ptotal-after = (2.8 kg) • 5 + (1.9 kg) • 5
ptotal-subsequently = (4.7 kg) • v
Bold an isolated organization, total arrangement momentum is conserved. Thus, before- and after-standoff momentum expressions tin can be fix equal to each other, and the equation can be manipulated to solve for the mail-collision speed of the two objects.
v = (126 kg • cm/s) / (4.7 kg)
v = 26.809 cm/s (~27 cm/southward)
The momentum change of the cart (Delta pcart) is merely the divergence betwixt the initial and final momentum values.
Delta pcart = mcart-after • vcart-subsequently - mcart • vcart-earlier
Delta pcart = (ii.8 kg) • (26.809 m/southward) - (2.8 kg) • (45 cm/s)
Delta pcart = -50.936 kg • cm/s (~51 kg•cm/s)
During the collision, the cart loses 50.9 units of momentum. Since full organization momentum is conserved, the brick must proceeds the aforementioned quantity of momentum.
The change in momentum of the cart is due to the fact that an impulse acts upon the cart during the collision. The moment contact is made betwixt the brick and the cart, the ii objects are moving at dissimilar speeds relative to each other. Consequently, in that location will be a friction strength interim between the 2 surfaces until the 2 objects maintain the same speed. That is, the brick volition pull astern upon the moving cart in order to slow it down; and the moving cart will pull frontward upon the stationary brick in gild to speed it up. These impulses are what cause the momentum changes. And these impulses are equal to the momentum changes. Thus,
Impulse on brick = Delta pbrick = +l.936 kg • cm/s (~51 kg•cm/s)
64. In a physics lab, a 0.500-kg cart moving at 36.four cm/s collides inelastically with a second cart which is initially at rest. The 2 carts move together with a speed of 21.viii cm/southward after the collision. Make up one's mind the mass of the second cart.
Answer: ~ 0.335 kg
This trouble involves a perfectly inelastic collision betwixt two carts. Thus, the post-collision velocity of the ii carts are identical. For communication sake, the carts will be referred to as Cart A and Cart B. The given data is:
The unknown to be solved for in this problem is the mass of Cart B (chiliadB).
The solution begins by setting the writing expressions for the full momentum of the organisation before and after the collision.
Before Collision: ptotal-before = (0.500 kg)•(36.4 cm/s) + (mB)•(0 cm/s)After Collision: ptotal-after = (0.500 kg)•(21.8 cm/southward) + (mB)•(21.8 cm/s)
Assuming momentum conservation, these expressions are ready equal to each other and and so algebraically manipulated to solve for the unknown (mB).
(0.500 kg)•(36.4 cm/s) = (0.500 kg)•(21.viii cm/southward) + (yardB)•(21.8 cm/s)
(0.500 kg)•(36.4 cm/s) - (0.500 kg)•(21.viii cm/due south) = (mB)•(21.8 cm/southward)
(7.30 kg•cm/due south) = (mB)•(21.8 cm/due south)
0.33486 kg = mB
mB = ~0.335 kg
65. A 9230-kg truck collides head on with a 1250-kg parked car. The vehicles entangle together and slide a linear altitude of 10.6 meters before coming to balance. Assuming a uniform coefficient of friction of 0.820 between the road surface and the vehicles, make up one's mind the pre-collision speed of the truck.
Answer: 14.8 one thousand/s
Here is an example of a more difficult problem involving the combination of momentum principles with information learned in other units of the course. The trouble involves a perfectly inelastic collision between a truck and a car. Thus, the post-standoff velocity of the truck and auto are identical. The given collision information is:
The unknown to be solved for in this problem is the velocity of the truck earlier the collision (fiveTruck-before).
At that place is some kinematic/dynamic data provided that volition assist in determining the post-collision velocity of the entangled truck and machine.
The combination of auto and truck will slide to a final resting position due to the action of friction. The coefficient of friction, a costless-body diagram and a kinematic equation tin be used to determine the velocity of the car and truck immediately following the standoff.
The complimentary-body diagram is shown at the right. Note that the unbalanced force is friction. Its value is found by multiplying the coefficient of friction by the combined weight of the car and truck (mu•Yard•thou where Thou = mtruck + mcar ). The strength of friction is the net force. Thus, the acceleration is the force of friction divided by the combined mass of the car and the truck (Thousand). Subsequently, the expression for the dispatch of the car and truck while sliding to a finish is but mu•g.
a = -8.036 m/s/s
(The - sign indicates a deceleration or slowing down move.)
Now a kinematic equation tin can be used to solve for the velocity of the car and truck immediately afterward the collision. This is shown below:
(0 one thousand/southward)2 = fiveo 2 + 2•(-8.036 m/s/s)•(10.six m) = 5o 2 - 170.36 one thousandtwo/s2
(0 k/s)2 = fiveo 2 - 170.36 grandii/due south2
5o two = 170.36 thousandtwo/s2
vo = SQRT(170.36 grandii/southii)
vo = 13.052 m/south (= vTruck-after = vMachine-after )
Now that the post-collision velocity of the machine and truck are known, expressions for the total system momentum tin can be written for the before- and subsequently-standoff situations.
Earlier Collision: ptotal-before = (9320 kg)•(vTruck-earlier) + (1250 kg)•(0 cm/s)After Collision: ptotal-later = (9320 kg)•(13.052 m/south) + (1250 kg)•(13.052 m/s)
Bold momentum conservation, these expressions are set equal to each other and then algebraically manipulated to solve for the unknown (thousandB).
(9320 kg)•(vTruck-before) = (9320 kg)•(13.052 m/due south) + (1250 kg)•(thirteen.052 m/s)
(9320 kg)•(5Truck-before) = 137963 kg•m/due south
(fiveTruck-before) = (137963 kg•m/s) / (9320 kg)
vTruck-before = ~fourteen.8 thousand/s
66. A classic physics demonstration involves firing a bullet into a cake of wood suspended by strings from the ceiling. The acme to which the forest rises below its everyman position is mathematically related to the pre-collision speed of the bullet. If a 9.7-gram bullet is fired into the center of a ane.one-kg cake of wood and it rises upward a distance of 33 cm, then what was the pre-collision speed of the bullet?
Answer: 2.9 x 10two m/southward
Here is another case in which momentum principles must be combined with content learned in other units in order to consummate an analysis of a concrete situation. The standoff involves the inelastic collision between a block of wood and bullet. The bulled lodges into the wood and the two objects move with identical velocity after the collision. The kinetic free energy of the wood and bullet is then converted to potential free energy as the combination of two objects rises to a terminal resting position.
Energy conservation tin can be used to make up one's mind the velocity of the wood-bullet combination immediately subsequently the collision. The kinetic energy of the wood-bullet combination is prepare equal to the last potential energy of the wood-bullet combination and the equation is manipulated to solve for the post-collision velocity of the forest-bullet combination. The work is shown here:
0.5 • fivecombination-after ii = (ix.8 one thousand/due south2) • (0.33 g)
fivecombination-later 2 = 2 •(9.8 m/s2) • (0.33 1000)
vcombination-after 2 = half dozen.468 m2/s2
5combination-later on = 2.5432 m/southward
At present momentum conservation can be used to determine the pre-collision velocity of the bullet (vbullet-before). The known information is:
Expressions for the total arrangement momentum can be written for the before- and later-collision situations.
Before Standoff: ptotal-before = (1.1 kg)•(0 m/s) + (0.0097 kg)•(vbullet-before)After Collision: ptotal-after = (1.ane kg)•(2.5432 g/due south) + (0.0097 kg)•(two.5432 m/south)
Assuming momentum conservation, these expressions are set equal to each other and then algebraically manipulated to solve for the unknown (thouB).
(0.0097 kg)•(vbullet-before) = (1.1 kg)•(ii.5432 m/s) + (0.0097 kg)•(2.5432 k/due south)
(0.0097 kg)•(5bullet-earlier) = 2.8222 kg•m/due south
vbullet-before = (2.8222 kg•k/south) / (0.0097 kg)
vbullet-before = 290.95 m/s
vbullet-before = ~ii.9 x 102 m/s
67. At an amusement park, twin brothers Timmy (thou = 50 kg) and Jimmy (chiliad = 62 kg) occupy separate 36-kg bumper cars. Timmy gets his automobile cruising at 3.six m/due south and collides head-on with Jimmy who is moving the opposite direction at 1.half dozen m/s. Afterward the collision, Timmy bounces backwards at 0.five m/s. Assuming an isolated system, determine ...
a. ... Jimmy'southward post-collision speed.b. ... the percent of original kinetic free energy which is lost as the consequence of the collision.
Reply: (a) v = ~two.0 one thousand/s
(b) % KE Loss = ~lxx. %
(a) Expressions for the total momentum of the organisation before and after the standoff can exist written. For the earlier-collision expression, Timmy is assigned a positive velocity value and Jimmy is assigned a negative velocity value (since he is moving in the opposite direction). Furthermore, the mass of the bumper car must exist figured into the full mass of the individually moving objects.
ptotal-earlier = one thousandTimmy • vTimmy-before + one thousandJimmy • 5Jimmy-earlier
ptotal-before = (86 kg) • (three.6 thousand/s) + (98 kg) • (-ane.half-dozen m/s)
For the earlier-standoff expression, Timmy is assigned a negative velocity value (since he has bounced backwards in the contrary direction of his original motion. Jimmy is assigned a velocity of v since his velocity is not known.
ptotal-after = mTimmy • vTimmy-after + mJimmy • vJimmy-after
pfull-after = (86 kg) • (-0.5 m/s) + (98 kg) • (vJimmy-after)
ptotal-after = (86 kg) • (-0.5 1000/due south) + (98 kg) • v
Since the system is assumed to exist isolated, the before-collision momentum expression can be fix equal to the after-standoff momentum expression. The equation can and so be algebraically manipulated to solve for the post-standoff velocity of Jimmy.
309.six kg•m/southward - 156.8 kg•m/s = (86 kg) • (-0.5 g/s) + (98 kg) • v
152.8 kg•m/s = - 43 kg•m/south + (98 kg) • v
152.eight kg•m/due south + 43 kg•yard/s = (98 kg) • v
195.8 kg•m/southward = (98 kg) • v
five = (195.8 kg•k/s) / (98 kg)
v = 1.998 m/s = ~2.0 m/south
(b) This collision is neither perfectly rubberband (since the collision force is a contact force) nor perfectly inelastic (since the objects do not stick together). Information technology is a partially elastic/inelastic standoff. Since the collision is non perfectly elastic, at that place is a loss of total system kinetic free energy during the collision. The before-standoff and subsequently-collision kinetic energy values can be calculated and the percent of total KE lost can be determined.
The earlier-collision KE is based on before-collision speeds:
KEarrangement-before = 0.5 • mTimmy • 5Timmy -before two + 0.5 • thousandJimmy • vJimmy-earlier 2
KEsystem-before = 0.5 • (86 kg) • (three.6 grand/s)2 + 0.5 • (98 kg) • (1.half dozen m/s)2
KEsystem-earlier = 557.28 J + 125.44 J
KEsystem-before = 682.72 J
The later-collision KE is based on later -collision speeds:
KEsystem-later on = 0.5 • thousandTimmy • vTimmy-after + 0.v • chiliadJimmy • vJimmy-after
KEsystem-subsequently = 0.5 • (86 kg) • (0.5 thou/s)2 + 0.five • (98 kg) • (1.998 m/south)2
KEsystem-later = ten.75 J + 195.61 J
KEsystem-after = 206.36 J
The system kinetic energy is changed from 682.72 J to 206.36 J during the collision. The full KE lost is 476.36 J. This value can exist used to make up one's mind the percent of the original KE which is lost in the collision.
% KE Loss = 69.77% = ~70%
68. Ii billiard balls, assumed to take identical mass, collide in a perfectly elastic standoff. Ball A is heading E at 12 m/s. Ball B is moving West at 8.0 thou/s. Make up one's mind the post-standoff velocities of Ball A and Ball B.
Reply: 5A-later on = -eight.0 cm/southward; vB-subsequently = 12 cm/s (The - indicates West and the + indicates East)
This collision is said to be perfectly elastic. Thus, both the total organisation momentum and the total system kinetic free energy of the ii objects is conserved. The momentum conservation equation can be written equally
Since the balls are identical, their masses are the same. That is, mA = chiliadB = m. The equation can exist rewritten as:
Since each term of the equation contains the variable grand, nosotros tin can separate through by grand and cancel chiliad'due south from the equation. The equation can be rewritten as:
vA-before + fiveB-before = vA-after + 5B-after For elastic collisions, total system kinetic energy is conserved. The kinetic energy conservation equation is written equally
As shown in the book, this equation tin can exist simplified to the grade of
fiveA-before + fiveA-after = 5B-before + fiveB-afterward The problem states the before-collision velocities of the 2 balls.
vB-before = - 8 cm/southward (the - indicates west)
These two values tin can be substituted into equations one and ii above.
12 cm/due south - eight cm/s = vA-after + vB-after 
12 cm/s + fiveA-after = - eight cm/south + vB-later
Now the problem has been reduced to two equations and two unknowns. Such a problem can be solved in numerous ways. One method involved using Equation three to develop an expression for vA-subsequently in terms of vB-after. This expression for vA-after can then be substituted into Equation 4. The value of vB-after tin so be adamant. This is shown below.
From Equation 3:
vA-later on = 4 cm/s - vB-after
This expression for 5A-after in terms of 5B-after can now be substituted into equation iv. This is shown below. The subsequent algebraic manipulation is shown equally well.
12 cm/s + 4 cm/s + viii cm/south - fiveB-after = + vB-afterwards
24 cm/due south = + 5B-subsequently + vB-after
24 cm/s = 2 vB-later on
5B-later = +12 cm/south
At present that the value of vB-subsequently has been adamant, it tin be substituted into the original expression for vA-after (Equation five) in order to determine the numerical value of vA-subsequently. This is shown beneath.
fiveA-after = 4 cm/s - 12 cm/s
5A-after = -8.0 cm/southward
69. A one.72-kg cake of soft forest is suspended by 2 strings from the ceiling. The forest is free to rotate in pendulum-like fashion when a forcefulness is exerted upon it. A 8.50-g bullet is fired into the wood. The bullet enters the wood at 431 one thousand/s and exits the contrary side shortly thereafter. If the woods rises to a tiptop of 13.eight cm, then what is the go out speed of the bullet?
Reply: vbullet-after = 98.2 m/due south
The difficulty of this trouble lies in the fact that information from other units (work and energy) must be combined with the momentum information from this unit to go far at a solution to the problem. In this scenario there is a collision between a stationary block of wood and a moving bullet. The impulse causes the block of woods to be gear up into motion and the bullet to slow downwards. Momentum can exist assumed to be conserved. One time set into motion, the block of forest rises in pendulum-similar fashion to a given height. Its energy of motility (kinetic energy) is transformed into energy of vertical position (potential energy). The mail-standoff speed of the forest tin can be determined using energy conservation equations.
To begin the solution, the final height of the wood is used to determine the mail service-collision speed of the woods.
0.5 • thousandwood •vwood ii = yardforest • g • hwood
vforest 2 = two • m • hwood
vwood = SQRT(2 • g • hwood)
vwood = SQRT[2 • (9.8 m/sii) • (0.138 m)]
vwood = SQRT[2.7048 thouii/south2]
vwood = 1.6446 1000/south
Immediately following the emergence of the bullet from the wood, the woods block is moving with a speed of ane.6446 m/south. Knowing this, momentum conservation can be applied to determine the post-collision speed of the bullet.
where vforest-earlier = 0 k/due south; vbullet-before = 431 1000/s; vwood-after = i.6446 yard/s; vbullet-later = ???
(ane.72 kg) • (0 m/due south) + (0.00850 kg) • (431 m/s) = (one.72 kg) • (1.6446 thou/due south) + (0.00850 kg) • vbullet-after
(To simplify the work, the units will be dropped from the solution in the next several steps. One time a 5bullet-after value is found, its units will exist in k/due south, consistent with the units stated in the above line.)
0 + 3.6635 = two.8288 + 0.00850 • vbullet-afterward
0.8347 = 0.00850 • 5bullet-after
(0.8347) / (0.00850) = vbullet-afterwards
5bullet-afterwards = 98.205 m/s = ~98.2 m/s
70. In a physics lab, the pitching speed of a pupil is determined by throwing a baseball game into a box and observing the box'due south motion after the grab. A measurement of the the distance the box slides across a rough surface of known coefficient of friction will allow one to determine the pre-impact speed of the pitched ball. If a 0.256-kg ball hits a 3.46-kg box and the ball and box slide a distance of ii.89 meters across a surface with a coefficient of friction of 0.419, then what is the pre-affect speed of the pitched ball?
Reply: 70.7 chiliad/s
This is some other example of a problem in which information from other units (work and energy or Newton'southward laws and kinematics) must be combined with the momentum information from this unit to get in at a solution to the trouble. There is a collision betwixt a stationary box and a moving baseball that causes the baseball to wearisome downward and the box to speed upwardly. It is a perfectly inelastic collision with the baseball remaining lodged in the box and the two objects moving together with the aforementioned mail-collision speed. After the collision occurs, the baseball and box slide a given distance across a rough surface to a final resting position. The coefficient of friction between the box and the surface is given. This latter information (sliding distance and mu value) can be used to decide the post-collision speed of the box and baseball. In one case plant, momentum conservation can exist applied to the collision to decide the pre-standoff speed of the baseball.
Work and free energy principles will be used to clarify the motion of the box/baseball game organization sliding to a stop. (Newton's laws and kinematics could just as hands been used). Immediately following the standoff, the box/baseball system has kinetic free energy. Friction does work upon the box/baseball system to bring to a terminal resting position, characterized by cypher kinetic free energy. The motility occurs across a level surface, and then there is no potential energy change of the box. The piece of work done past friction is equal to the kinetic energy modify of the box/baseball game system.
Ffrict • d • cos(Theta) = KEfinal - KEinitial
(mu •Fnorm) • d • cos(180) = 0.5 • chiliad •vfinal two - 0.5 • 1000 •vinitial two
- (mu • m • thousand) • d = 0 - 0.5 • one thousand • 5initial two
mu • chiliad • d = 0.5 • 5initial 2
2 •mu • g • d = vinitial 2
SQRT(two • mu • g • d) = vinitial
where mu = 0.419; d = two.89 m; and grand = ix.8 m/s2
vinitial = SQRT[ii • (0.419) • (9.8 m/s2 ) • (2.89 m)]
vinitial = SQRT[23.734 mtwo/s2]
vinitial = 4.8717 one thousand/s
Immediately post-obit the collision, the box/baseball game arrangement begins moving with a speed of 4.8717 m/s. Now momentum conservation tin can be applied to determine the pre-standoff speed of the baseball.
(0.256 kg) • vbaseball-before + (3.46 kg) • (0 grand/due south) = (0.256 kg) • (4.8717 1000/south) + (3.46 kg) • (4.8717 m/s)
(To simplify the solution, the units volition exist dropped from the solution in the next several steps. Once a 5baseball-before value is found, its units will be in m/s, consequent with the units stated in the above line.)
0.256 • fivebaseball-before = 1.2472 + 16.8562
0.256 • vbaseball game-earlier = 18.1033
fivebaseball-before = (18.1033) / (0.256)
vbaseball game-before = 70.vii one thousand/southward
71. Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1-kg skater moving Eastward at 4.33 chiliad/s. The two skaters entangle and move together across the ice. Determine the magnitude and direction of their post-collision velocity.
Respond: 3.64 m/s at 42.5 degrees e of s (312.v degrees)
The difficulty of this problem lies in the fact that the collision occurs between ii objects moving at right angles to each other. Thus, vector principles will have to be combined with momentum principles to arrive at a solution to the trouble. The same conservation of momentum principle volition exist used; but when summing the earlier momentum values of the two objects, the fact that they are at right angles to each means that they volition accept to be added using the Pythagorean theorem. The collision is perfectly inelastic with the two skaters moving at the aforementioned speed afterwards the standoff. For communication sake, the 39.6-kg skater will be referred to as skater A and the 52.1-kg skater will be referred to every bit skater B. A vector diagram will probable assistance in the solution of the problem.
The individual momentum of the two skaters is first determined.
pB = mB • B = (52.1 kg) • (4.33 chiliad/due south, East ) = 225.593 kg • thou/southward, E
Now the Pythagorean theorem can exist used to add these 2 vectors and thus decide the pre-collision organisation momentum. The diagram at the right shows the vectors being added in caput-to-tail style. The resultant is drawn from the tail of the first vector to the head of the last vector. The resultant is the hypotenuse of a correct triangle whose sides are the pA and pB vectors.
porganization = pA + pB (where pA and pB are correct angle vectors) psystem = SQRT(pA ii+ pB 2)
psystem = SQRT[(245.916 kg • m/southward)2+ (225.593 kg • m/s)2]
psystem = 333.717 kg • m/s
The management of the this total system momentum vector can be determined by using a trigonometric function. As shown in the diagram above, the angle theta is the angle between the system momentum vector and the vertical. This angle can be determined using either the tangent, cosine or sine function. The tangent function is used below.
tangent(Theta) = pB / pA
tangent(Theta) = (225.593 kg • grand/southward) / (245.916 kg • m/southward)
tangent(Theta) = 0.91735
Theta = tan-1 (0.91735)
Theta = 42.532 degrees
Before the standoff, the total organisation momentum is 333.717 kg • g/s in a direction of 42.532 degrees east of south. Since total system momentum is conserved, the after-collision momentum of the system is also 333.717 kg • m/southward in a direction of 42.532 degrees east of due south. After the collision, the ii objects movement together as a single unit of measurement with the aforementioned velocity. The velocity of each object tin be found by dividing the total momentum by the total mass.
vsystem = ( psystem ) / (msystem )
vsystem = (333.717 kg • thou/due south) / (91.7 kg)
vsystem = 3.64 m/s at 42.5 degrees east of south
72. In a physics lab, two carts collide elastically on a level, low-friction track. Cart A has a mass of i.500 kg and is moving due east at 36.5 cm/southward. Cart B has a mass of 0.500 kg and is moving West at 42.8 cm/s. Decide the postal service-collision velocities of the two carts.
Respond: 5A-later = -3.fifteen cm/s; vB-after = 76.15 cm/due south
This is a perfectly elastic collision in which both momentum and kinetic energy are conserved. The method for solving this problem will exist very like to that used in Problem #68 above. Ii equations will exist developed using the momentum conservation and kinetic energy conservation principles. One equation will be used to develop an expression for vA in terms of vB. This expression will then be substituted into the second equation in order to solve for vB. The original vA expression tin then be used to determine the vA value. The solution is shown below.
The momentum conservation equation tin be written as
(1.500 kg) • (+36.v cm/s) + (0.500 kg) • (-42.8 cm/s) = (i.500 kg) • 5A-later on + (0.500 kg) • vB-after
33.35 kg • cm/s = (1.500 kg) • vA-afterwards + (0.500 kg) • 5B-afterwards
For elastic collisions, total system kinetic free energy is conserved. The kinetic energy conservation equation is written equally
As shown in the volume, this equation can be simplified to the class of
36.5 cm/s + vA-afterward = -42.8 cm/due south + vB-subsequently
vA-after = vB-afterwards - 79.3 cm/s
Now the problem has been reduced to 2 equations and ii unknowns. Such a problem can exist solved in numerous ways. Note that equation ii represents an expression for vA-afterward in terms of vB-after. This expression for vA-later on tin then exist substituted into Equation ane. The value of vB-after can then be determined. This piece of work is shown beneath. (To simplify the mathematics, the units will be dropped from the numerical values stated in the solution. When vB-afterwards is solved for, its units will be in cm/s - the aforementioned units used for velocity in the in a higher place portion of the solution.)
33.35 = 1.500 • vB-after - 118.95 + 0.500 • vB-later on
152.30 = two.00 • 5B-after
5B-after = 76.xv cm/southward
Now that the value of 5B-after has been adamant, it can be substituted into the original expression for vA-subsequently (Equation 2) in society to determine the numerical value of vA-later on. This is shown below.
vA-after = 76.15 cm/s - 79.3 cm/due south
vA-afterwards = -iii.15 cm/southward
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