# Calculating Work Done during a Collision

• naianator
In summary, the conversation discussed finding the work done by the contact force on block m2 during a collision with block m1. The correct equation for this is W = 1/2*m_2*((m_1*v_1-m_2*v_2)/m_2)^2 - 1/2*m_2*v_2^2, and the work done on mass m1 by mass m2 can be calculated as -1/2*m_1*v_1^2.
naianator

## Homework Statement

Two blocks have masses m1 and m2 and speeds v1 and v2. The objects slide directly toward each other along a frictionless horizontal surface and collide. After the collision, block m1 is at rest. Find the work done by the contact force on block m2 during the collision.

## Homework Equations

W = 1/2*m*v_f^2 - 1/2*m*v_i^2

m*v_1 + m*v_2 = m*v_1' + m*v_2'

## The Attempt at a Solution

m_1*v_1+m_2*v_2 = 0+m_2v_2'

v_2' = (m_1*v_1+m_2*v_2)/m_2

W = 1/2*m_2*((m_1*v_1+m_2*v_2)/m_2)^2 - 1/2*m_2*v_2^2

What am I doing wrong?

Post edited .

Last edited:
m_1*v_1+m_2*v_2 = 0+m_2v_2' is wrong all v-s are speeds to be corrected as
m_1*v_1- m_2*v_2 = 0+m_2v_2' ----------------------------- (1), hence
v_2' = (m_1*v_1-m_2*v_2)/m_2 or
W(by m1 on m2) = 1/2*m_2*((m_1*v_1-m_2*v_2)/m_2)^2 - 1/2*m_2*v_2^2, simplify to get the correct answer.

Work done on mass m1 by mass m2 = -1/2*m_1*v_1^2

Let'sthink said:
m_1*v_1+m_2*v_2 = 0+m_2v_2' is wrong all v-s are speeds to be corrected as
m_1*v_1- m_2*v_2 = 0+m_2v_2' ----------------------------- (1), hence
v_2' = (m_1*v_1-m_2*v_2)/m_2 or
W(by m1 on m2) = 1/2*m_2*((m_1*v_1-m_2*v_2)/m_2)^2 - 1/2*m_2*v_2^2, simplify to get the correct answer.

Work done on mass m1 by mass m2 = -1/2*m_1*v_1^2
ahhh yes, thank you!

## 1. What is work done during collision?

Work done during collision refers to the amount of energy transferred or transformed during a collision between two objects. It is a measure of the force exerted by one object on another during a collision.

## 2. How is work done during collision calculated?

The work done during collision can be calculated using the formula W = Fd, where W is the work done, F is the force applied, and d is the distance the object moves in the direction of the force.

## 3. Is work done during collision always positive?

No, work done during collision can be either positive or negative. If the force and displacement are in the same direction, the work done is positive. If the force and displacement are in opposite directions, the work done is negative.

## 4. What factors affect the work done during collision?

The factors that affect the work done during collision include the mass and velocity of the objects involved, the angle of collision, and the type of collision (elastic or inelastic).

## 5. How is work done during collision related to kinetic energy?

According to the work-energy theorem, the work done during collision is equal to the change in kinetic energy of the objects involved. This means that the work done during collision can either increase or decrease the kinetic energy of the objects.

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