# Work done during a collision -- Change in Kinetic Energy & change in Momentum

• Aerisk16
Aerisk16
Homework Statement
work done in an collision which lasted for a very brief time interval
Relevant Equations
J=∆p=p_2-p_1=mv_2-mv_1
J=∑F∆t=F_ave ∆t=F_ave (t_2-t_1 )
W=Fs=∆K=K_2-K_1=1/2 mv_2-1/2 mv_1
Hello guys,

I need help on this problem,

"You throw a ball with a mass of 0.4kg against a brick wall. It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20m/s. (a) Find the Impulse of the net force on the ball during its collision with the wall (b) If the ball is in contact with the wall for 0.010s, find the average horizontal force that the wall exerts on the ball during the impact."

in the book, the answer for the Impulse is 20 N.s (there's a change in momentum) and the average force is 2000 N determine by dividing the Impulse 20 N.s by the time interval 0.01 s. Since there is change in velocity or speed, there is a change in kinetic energy and since the total work is equal to change in kinetic energy, I divided the change in kinetic energy which is -100 J by the average force 2000 N and get -0.05 m or 5 cm as displacement, I'm thinking this is a deformation of the ball. I'm having a doubt with my answer and also the size of the ball is not given...

berkeman
Aerisk16 said:
divided the change in kinetic energy which is -100 J by the average force 2000 N and get -0.05 m or 5 cm as displacement, I'm thinking this is a deformation of the ball. I'm having a doubt with my answer and also the size of the ball is not given...
Your thinking is good. The conclusion may not be as solid as you hope.

Under the assumption of a constant force of 2000 N there must be a 5 cm difference between the compression and the rebound. That would account for a permanent 5 cm deflection of the ball's surface, just as you conclude.

Let us pause for a quick sanity check. The momentum lost and the momentum regained are in a 3:2 proportion. If force is constant then the compression takes 0.006 seconds and rebound takes 0.004 seconds.

That is 0.006 seconds times an average speed of 30 m/s divided by two. That is a 9 cm compression
That is 0.004 seconds times an average speed of 20 m/s divided by two. That is a 4 cm rebound.

Net 5 cm deflection as expected. Sanity check passes.

So we have a ball with a minimum diameter of 9 cm that now has a 5 cm deep dent.But we do not know that force is constant. Indeed, we would expect otherwise. There might, for instance be a force greater than 2000 N during compression and a force less than 2000 N during the rebound. The average force could still be 2000 N and the deflection during compression and rebound could be identical.

Like a slightly under-inflated basketball bouncing from a wall but retaining its original size and shape.It may be worth noting that the momentum analysis allows you to deduce the time-weighted average force. But for an energy analysis, you need a displacement-weighted force. The two averages need not match.

PeroK

## What is the relationship between work done during a collision and the change in kinetic energy?

The work done during a collision is directly related to the change in kinetic energy. According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. In a collision, the forces involved do work on the colliding bodies, leading to changes in their kinetic energies.

## How does momentum change during a collision?

Momentum changes during a collision depending on the type of collision and the forces involved. In an elastic collision, total momentum is conserved and redistributed among the colliding bodies. In an inelastic collision, momentum is still conserved, but some of it may be converted into other forms of energy, such as heat or deformation.

## Is it possible for kinetic energy to be conserved in a collision?

Yes, kinetic energy can be conserved in a collision, but only in an elastic collision. In an elastic collision, both kinetic energy and momentum are conserved. In inelastic collisions, kinetic energy is not conserved because some of it is converted into other forms of energy, although momentum is still conserved.

## How can we calculate the work done during a collision?

The work done during a collision can be calculated by finding the change in kinetic energy of the system. This can be done using the formula: Work = ΔKE = KE_final - KE_initial, where KE represents the kinetic energy of the system before and after the collision.

## What factors affect the change in momentum during a collision?

The change in momentum during a collision is affected by the masses of the colliding bodies, their initial velocities, and the nature of the collision (elastic or inelastic). External forces, if present, can also affect the total momentum of the system. In a closed system with no external forces, the total momentum before and after the collision remains constant.

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