Question

A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.

What is the velocity after the collision?

m/s

Answer 1

The velocity after the collision is 2.82 m/s

Step-by-step explanation:

Law Of Conservation Of Linear Momentum

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

`P=m_1v_1+m_2v_2`

If a collision occurs and the velocities change to v', the final momentum is:

`P'=m_1v'_1+m_2v'_2`

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

`m_1v_1+m_2v_2=m_1v'_1+m_2v'_2`

If both masses stick together after the collision at a common speed v', then:

`m_1v_1+m_2v_2=(m_1+m_2)v'`

The common velocity after this situation is:

`\displaystyle v'=(m_1v_1+m_2v_2)/(m_1+m_2)`

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

`\displaystyle v'=(3.91*5.7+4*0)/(3.91+4)`

`\displaystyle v'=(22.287)/(7.91)`

`\boxed{v' = 2.82\ m/s}`

The velocity after the collision is 2.82 m/s