# Elastic and Inelastic collisions assignement problems. 1st year physics

• torth76
In summary, the first problem involves two inelastic balls colliding with velocities of -4m/s and 7m/s. The solution involves using the equations m1v1 + m2v2 = m1V1 + m2V2 and v1 + V1 + v2 + V2, and after substitution and factoring, the final velocity of one of the balls is found to be equal to the initial velocity. The second problem involves a 0.5kg mass of silly putty and a 0.003kg paintball pellet colliding inelastically 2.18m below the top of a 100m tall building. The solution involves using kinematics to find the velocities at impact and
torth76
Here are 3 problems I have been working on. Any help would be very much appreciated. Thank you in advance!

## Homework Statement

Two inelastic balls are traveling toward each other with velocities -4m/s and 7 m/s and they experience an elastic head on collision
a)obtain the velocities (magnitude and direction) of each ball after the collision
b) recalculate the results for the ball moving at 7m/s being 2x the mass of the other ball

## Homework Equations

m1v1 + m2v2 = m1V1 + m2V2
v1 + V1 + v2 + V2

where v= initial and V=final

## The Attempt at a Solution

i feel to solve this i need to substitute one of the equations into the other and solve for one of the final velocities.

I did this and obtained

m1v1 + m2v2 - m1v2 + m1v1 = V (m1 +m2)

since masses are equal i factored like this

m(v1 +v2 -v2 +v1)/2m = V
m(v1+v1)/2m = V
mv1(2)/2m = V
v1=V

I feel like this is wrong. I have done it a few times and get different answers. I don't account for the direction of ball 2 until i plug in a value as -4, until then I leave it as positive.

## Homework Statement

A 0.5kg mass of silly putty is at rest at the top of a 100m tall building; it is dropped. At the same time a 0.003kg paintball pellet is fight straight up towards the ground with an initial veloctiy of 150 m/s. The paintball and putty collide inelastically 2.18m below the top of the building. Measured from the ground what is the maximun height that the combined putty/paintball mass achieves?

## Homework Equations

v^2 = v0^2 +2ax

m1v01 + m2v02 = (m1 +m2)Vf

## The Attempt at a Solution

i first used kinematics to find the velocities of the objects right at impact. the putty = -6.54m/s and the pellet =143.46.

Next I used m1v01 + m2v02 = (m1 +m2)Vf to solve for the velocity of the combined mass right after the collision. I found it to be -5.65m/s. I am not sure why my answer is negative and can't see why but I know it should be a positive value. If i use the positive value and use kinematics to solve for the height once it decelerates to v=0m/s i find it goes 1.63m equaling 99.45m above the ground.

I am unsure of this answer. I feel like my method is right but I am not sure if what I did was correct.

## Homework Statement

1 0.5kg mass of silly putty in suspended 5m above the floor by a long string. A paint ball pellet (m=0.003kg) is fired at the putty from the floor with an initial velocity of 150m/s. the two masses collide inelastically. what is the minimum length of the string such that the combined putty/paint ball mass does not splatter into the ceiling?

## Homework Equations

v^2 = v0^2 + 2ax
m1v01 +m2v02 = )m1+m2)Vf

## The Attempt at a Solution

using kinematics i determined the speed of the pellet before impact would be 149.67m/s. Using the equation for inelastic collision i found that the v0 of the ball is 0 and the vf of the combined mass to be 0.89m/s. Using this value I used the kinematics equation again and found distance (x) it would go until it came to rest. I found this value to be 0.04m.

After working through this problem I don't see what I am doing wrong. I don't think my answer makes sense since the string would only have to be longer than 4cm.

torth76 said:
Here are 3 problems I have been working on. Any help would be very much appreciated. Thank you in advance!

## Homework Statement

Two inelastic balls are traveling toward each other with velocities -4m/s and 7 m/s and they experience an elastic head on collision
a)obtain the velocities (magnitude and direction) of each ball after the collision
b) recalculate the results for the ball moving at 7m/s being 2x the mass of the other ball

## Homework Equations

m1v1 + m2v2 = m1V1 + m2V2
v1 + V1 + v2 + V2

where v= initial and V=final

## The Attempt at a Solution

i feel to solve this i need to substitute one of the equations into the other and solve for one of the final velocities.

I did this and obtained

m1v1 + m2v2 - m1v2 + m1v1 = V (m1 +m2)

since masses are equal i factored like this

m(v1 +v2 -v2 +v1)/2m = V
m(v1+v1)/2m = V
mv1(2)/2m = V
v1=V

I feel like this is wrong. I have done it a few times and get different answers. I don't account for the direction of ball 2 until i plug in a value as -4, until then I leave it as positive.

## Homework Statement

A 0.5kg mass of silly putty is at rest at the top of a 100m tall building; it is dropped. At the same time a 0.003kg paintball pellet is fight straight up towards the ground with an initial veloctiy of 150 m/s. The paintball and putty collide inelastically 2.18m below the top of the building. Measured from the ground what is the maximun height that the combined putty/paintball mass achieves?

## Homework Equations

v^2 = v0^2 +2ax

m1v01 + m2v02 = (m1 +m2)Vf

## The Attempt at a Solution

i first used kinematics to find the velocities of the objects right at impact. the putty = -6.54m/s and the pellet =143.46.

Next I used m1v01 + m2v02 = (m1 +m2)Vf to solve for the velocity of the combined mass right after the collision. I found it to be -5.65m/s. I am not sure why my answer is negative and can't see why but I know it should be a positive value. If i use the positive value and use kinematics to solve for the height once it decelerates to v=0m/s i find it goes 1.63m equaling 99.45m above the ground.

I am unsure of this answer. I feel like my method is right but I am not sure if what I did was correct.

## Homework Statement

1 0.5kg mass of silly putty in suspended 5m above the floor by a long string. A paint ball pellet (m=0.003kg) is fired at the putty from the floor with an initial velocity of 150m/s. the two masses collide inelastically. what is the minimum length of the string such that the combined putty/paint ball mass does not splatter into the ceiling?

## Homework Equations

v^2 = v0^2 + 2ax
m1v01 +m2v02 = )m1+m2)Vf

## The Attempt at a Solution

using kinematics i determined the speed of the pellet before impact would be 149.67m/s. Using the equation for inelastic collision i found that the v0 of the ball is 0 and the vf of the combined mass to be 0.89m/s. Using this value I used the kinematics equation again and found distance (x) it would go until it came to rest. I found this value to be 0.04m.

After working through this problem I don't see what I am doing wrong. I don't think my answer makes sense since the string would only have to be longer than 4cm.

Your answer to part 3 is correct. The paint ball pellet has almost negligible mass compared to the 0.5 kg mass, so the pair will move up very little.

In Part 2, your answer of -5.65 is correct. The falling mass will be slowed slightly, but not stopped, so the maximum height reahed by the pair is the point of impact.

For Q1, try calculating the velocity of the centre of mass. Each ball will be approaching the centre of mass at a certain speed. They will be leaving the centre of mass at those same speeds after collision. Even when the masses are different.

## 1. What is the difference between elastic and inelastic collisions?

Elastic collisions are collisions where both the kinetic energy and momentum are conserved, meaning the total energy of the system before and after the collision remains the same. Inelastic collisions, on the other hand, do not conserve kinetic energy, meaning some of the energy is lost in the form of heat or sound. Momentum, however, is still conserved in inelastic collisions.

## 2. How do you calculate the coefficient of restitution for an elastic collision?

The coefficient of restitution is the ratio of the relative velocity of separation to the relative velocity of approach. In an elastic collision, this value is always 1, as the objects bounce off each other with the same speed they had before the collision.

## 3. What is the equation for conservation of momentum in an inelastic collision?

The equation for conservation of momentum in an inelastic collision is m1v1 + m2v2 = (m1 + m2)v, where m1 and m2 are the masses of the colliding objects, v1 and v2 are their velocities before the collision, and v is their velocity after the collision.

## 4. Can the total kinetic energy of a system increase in an inelastic collision?

No, the total kinetic energy of a system cannot increase in an inelastic collision. Some of the initial kinetic energy is always lost in the form of heat or sound, resulting in a decrease in total kinetic energy.

## 5. How do you determine whether a collision is elastic or inelastic?

To determine if a collision is elastic or inelastic, you can calculate the coefficient of restitution using the relative velocities of the colliding objects before and after the collision. If the coefficient of restitution is equal to 1, the collision is elastic, and if it is less than 1, the collision is inelastic.

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