Question

A 5.00-kg bullet moving with an initial speed of V_i= 400 m/s is fired into and passes through a 1.00-kg block as shown in the Figure. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 1000 N/m. The block moves d = 5.00 cm to the right after impact before being brought to rest by the spring. Find:

The speed at which the bullet emerges from the block.

The kinetic energy lost (the amount of intial kinetic energy of the bullet that is converted into internal energy, i.e. heat, in the bullet-block system) during the collison.

Answer 1

Given is:-

mass of the bullet = 0.005kg

Initial speed of the bullet = 400m/s

Mass of the block = 1 kg

Spring constant = 1000N/m

Distance moved by the block = 0.05 m

Now,

Since there are no external forces acting in the x direction on the system (bullet and block), momentum is conserved.

or

We need to find VBf in order to solve the problem. We can find VBf by considering . Immediately following the impact:

but V_{B final} = 0 and V_{B initial} = V_Bf and

we get

by plugging all the values we get

thus

Hence the speed at which the bullet emerges from the block is 84m/s

Part-b

by plugging all the values we get

This is the the amount of kinetic energy lost in the form of heat