A 57 kg basketball player jumps vertically upward to make a shot. At the very top of her jump, she throws the basketball with a speed of 4.4 m/s. What is her recoil velocity immediately after she releases the ball? The mass of the basketball is 0.62 kg.
In this problem, you are asked to find the recoil velocity of a basketball player after the ball is thrown. Any time you are asked to find speed or velocity of an object just before or just after a collision or separation, you should check to see if you can use Conservation of Momentum to solve the problem.
In this case, you are asked only for recoil velocity and so you know you don’t need to include effects of gravity or other external forces on the player. You can, indeed, use Conservation of Momentum.
For Conservation of Momentum problems, you always draw a picture of the system immediately before the collision or separation and another picture immediately after. Because momentum depends on mass and velocity, label all mass and velocity information
P_{b} = P_{a}
Any time you understand the motion of a system for which F_{extermal} Δt≈0, you begin with the Conservation of Momentum equation.
The basketball player’s recoil velocity is 0.048 m/s opposite to the direction at which the ball was shot. No further mathematical solution is needed for this problem.
It is certainly true that both the player and the ball feel the strength of the force between them equally. So you can use the motion of the basketball to solve for the force (in terms of Δt which is unknown) and put it into the Second Law equation for the player (the time intervals will divide out.) However, the equation that you develop as you do so is just the Conservation of Momentum equation.
In this problem, you were very specifically asked about recoil velocity of the player, or the velocity due to the backwards push of the ball only. It is true the velocity of the player will change from that value as gravity pulls her back down, so you cannot use Conservation of Momentum alone to find her speed as she hits the ground.
In general when you have a system that separates into pieces over a very short time interval, the internal force (causing the separation) is greater than external forces on the system. Therefore, as long as you compare the point in time immediately before the separation to the point in time immediately after the separation, the impulse by external forces (F_{extermal} Δt) is small enough that you can take momentum to be conserved.
The quick answer is that whenever you have a collision between objects, or a separation of one object into smaller objects, you can probably treat the momentum of that multiple object system as being conserved around the time of the collision/separation. The momentum of each individual object, however, changes.
The reason for this is that momentum is conserved on a system only when the net force on that system is small enough that F_{extermal} Δt is essentially zero. (This comes straight from Newton’s Second Law and the definition of acceleration.) In the case of a collision or separation, the force between the interacting objects or pieces is quite large. So the momentum of each individual object changes because of the force between them. (In this case, the velocity of the basketball and the velocity of the player each change as the ball is thrown.) But the net effect of the internal force on the two object system is zero (think of Newton’s Third Law.) So if we treat the interacting objects together as a single system, momentum of that larger system only changes by F_{extermal} Δt, or about zero. Δt is the time over which the collision or separation takes place.
The basketball player jumps straight up—there is no x-component to her jump. At the top of her motion, her y-velocity is also zero. Therefore, immediately before she shoots, the velocity of the player and the ball is zero.
Before the ball is thrown, the player and ball are moving together. Therefore, I chose to treat them as a single object with a combined mass of 57.62 kg. It is equally fine to treat them in two terms—velocity will be the same for both.
The final momentum is the (vector) sum of the moment a of each of the pieces—the ball and the player. For a vector equation, put in the correct sign for values that you know (such as the velocity of the ball) and your answer will then give you the correct sign for the unknown quantity (the velocity of the player.)
The problem states that the basketball player throws the ball—in other words, the “player-ball” system separates into two pieces. This is a candidate for a Conservation of Momentum problem.
Because the problem only asks for recoil velocity, you know you only need to consider the effects of the force of the throw—an internal force. So the conditions for using Conservation of Momentum are met.
Collisions or separations do not need to be dramatic. You can use Conservation of Momentum on any collision/separation problem in which F_{extermal} Δt≈0 over the time of the interaction. In other words, the internal force of interaction just needs to be significantly larger than the external forces on the system.
Momentum is conserved on a system only when the net force on that system is small enough that F_{extermal} Δt is essentially zero. (This comes straight from Newton’s Second Law and the definition of acceleration.) In the case of a collision or separation, the force between the interacting objects or pieces is quite large. So the momentum of each individual object changes because of the force between them. (In this case, the velocity of the basketball and the velocity of the player each change as the ball is thrown.) But the net effect of the internal force on the two object system is zero (think of Newton’s Third Law.)
In this case, the problem only asks for recoil velocity. In other words, you only need to consider the effects of the force of the throw—an internal force. So the conditions for using Conservation of Momentum are met.
The basketball player jumps straight up—there is no x-component to her jump. At the top of her motion, her y-velocity is also zero. Therefore, immediately before she shoots, the velocity of the player holding the ball is zero.
Before the separation, the player and ball are together and so the mass of the initial object is their combined mass.
The recoil velocity results from the force of the throw. Therefore, you need to compare the momentum of the system immediately before the throw to that immediately after. If you look over a longer time interval, you will not be able to ignore the effects of gravity and so cannot use Conservation of Momentum.
I use the subscript “b” to refer to the system at the time just before the throw, and “a” to refer to the system just after. You should use whatever subscripts are helpful for you to keep track of the player, the ball, and the two times.
We are not told the direction at which the player shoots the ball. However, we do know that momentum is conserved and so the recoil velocity will be opposite to the ball’s velocity.
Whenever you have a collision between objects, or a separation of one object into smaller objects, you can probably treat the momentum of that multiple object system as being conserved around the time of the collision/separation. The momentum of each individual object, however, changes. Therefore, you need to treat the objects together as your system.
The reason for this is that momentum is conserved on a system only when the net force on that system is small enough that F_{extermal} Δt is essentially zero. (This comes straight from Newton’s Second Law and the definition of acceleration.) In the case of a collision or separation, the force between the interacting objects or pieces is quite large. So the momentum of each individual object changes because of the force between them. (In this case, the velocity of the basketball and the velocity of the player each change as the ball is thrown.) But the net effect of the internal force on the two object system is zero (think of Newton’s Third Law.) So if we treat the interacting objects together as a single system, momentum of that larger system only changes by F_{extermal} Δt, or about zero.
This condition is usually met for collision or separation problems.
The values before the throw (b) can be put into the equation directly from your picture.
The values after the throw (a) can be put into the equation directly from your picture.
Momentum is a vector and so you need to treat components separately. Because the motions considered in this problem are along a single direction, I chose that to be my x- axis.
Before the ball is thrown, the player and ball are moving together. Therefore, I chose to treat them as a single object with a combined mass of 57.62 kg. It is equally fine to treat them in two terms—velocity will be the same for both.
The final momentum is the (vector) sum of the moment a of each of the pieces—the ball and the player. For a vector equation, put in the correct sign for values that you know (such as the velocity of the ball) and your answer will then give you the correct sign for the unknown quantity (the velocity of the player.)
The basketball player jumps straight up—there is no x-component to her jump. At the top of her motion, her y-velocity is also zero. Therefore, immediately before she shoots, the velocity of the player and the ball is zero.
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
In this problem, you were very specifically asked about recoil velocity of the player, or the velocity due to the backwards push of the ball only. It is true the velocity of the player will change from that value as gravity pulls her back down, so you cannot use Conservation of Momentum alone to find her speed as she hits the ground.
In general when you have a system that separates into pieces over a very short time interval, the internal force (causing the separation) is greater than external forces on the system. Therefore, as long as you compare the point in time immediately before the separation to the point in time immediately after the separation, the impulse by external forces (F_{extermal} Δt) is small enough that you can take momentum to be conserved.
The quick answer is that whenever you have a collision between objects, or a separation of one object into smaller objects, you can probably treat the momentum of that multiple object system as being conserved around the time of the collision/separation. The momentum of each individual object, however, changes.
The reason for this is that momentum is conserved on a system only when the net force on that system is small enough that F_{extermal} Δt is essentially zero. (This comes straight from Newton’s Second Law and the definition of acceleration.) In the case of a collision or separation, the force between the interacting objects or pieces is quite large. So the momentum of each individual object changes because of the force between them. (In this case, the velocity of the basketball and the velocity of the player each change as the ball is thrown.) But the net effect of the internal force on the two object system is zero (think of Newton’s Third Law.) So if we treat the interacting objects together as a single system, momentum of that larger system only changes by F_{extermal} Δt, or about zero. Δt is the time over which the collision or separation takes place.
It is certainly true that both the player and the ball feel the strength of the force between them equally. So you can use the motion of the basketball to solve for the force (in terms of Δt which is unknown) and put it into the Second Law equation for the player (the time intervals will divide out.) However, the equation that you develop as you do so is just the Conservation of Momentum equation.
In this problem, you are asked to find recoil velocity of a basketball player as she shoots a ball. This is a separation problem—a larger object separates into two pieces—and so you should first consider solving it using Conservation of Momentum. Because the problem only asked for recoil velocity (the effect of the force of the throw,) you are able to ignore any effect of external forces and so conditions to use Conservation of Momentum are, indeed, met.
Physically, you should expect the recoil velocity of the person to be small. Her mass is approximately 100 times that of the ball, and so if the magnitude of momentum for each is to be equal, her velocity should be about 1/100th that of the ball.