Question

A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2 m/s in the opposite direction on a horizontal frictionless surface. If the block is brought to rest by the collision, what is the kinetic energy of the bullet as it emerges from the block

Answer 1

`K=512J`

Step-by-step explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass
`m_1` has an initial velocity
`v_(1i)` and a final velocity
`v_(1f)` and the block of mass
`m_2` has an initial velocity
`v_(2i)` and a final velocity
`v_(2f)` then the initial and final momentum of the system will be:

`p_i=m_1v_(1i)+m_2v_(2i)`

`p_f=m_1v_(1f)+m_2v_(2f)`

Since momentum is conserved,
`p_i=p_f`, which means:

`m_1v_(1i)+m_2v_(2i)=m_1v_(1f)+m_2v_(2f)`

We know that the block is brought to rest by the collision, which means
`v_(2f)=0m/s` and leaves us with:

`m_1v_(1i)+m_2v_(2i)=m_1v_(1f)`

which is the same as:

`v_(1f)=(m_1v_(1i)+m_2v_(2i))/(m_1)`

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

`v_(1f)=((0.01kg)(2000m/s)+(4kg)(-4.2m/s))/(0.01kg)=320m/s`

So kinetic energy of the bullet as it emerges from the block will be:

`K=(mv^2)/(2)=((0.01kg)(320m/s)^2)/(2)=512J`