Question

A 0.01 kg bullet is fired at a 0.5 kg block initially at rest. The bullet, moving with an initial speed of 400 m/s, emerges from the block with a speed of 300 m/s. What is the speed of the block after the collision?

Answer 1

The speed of the block after the collision is 1 m/s.

Step-by-step explanation:

To find the speed of the block after the collision, we can use the principle of conservation of momentum. Before the collision, the total momentum of the system (bullet + block) is zero since the block is initially at rest. After the collision, the total momentum of the system is still zero.

Since momentum is conserved, we can write:

mass of bullet * initial velocity of bullet + mass of block * 0 = mass of bullet * final velocity of bullet + mass of block * final velocity of block

Plugging in the given values, we have: 0.01 kg * 400 m/s + 0.5 kg * 0 = 0.01 kg * 300 m/s + 0.5 kg * final velocity of block

Simplifying the equation, we find: 4 + 0 = 3 + 0.5 * final velocity of block

Solving for the final velocity of the block, we get: final velocity of block = (4 - 3) / 0.5 = 1 m/s

Therefore, the speed of the block after the collision is 1 m/s.

Answer 2

Use the law of conservation of momentum to solve this problem. We have a system of two bodies (bullet and block). Initially, only the bullet has a non-zero momentum. After the collision, both have some momentum and we know the part for the bullet, so it is simple to isolate the part for the block. Call v_t0 the initial bullet speed, v_t1 new bullet speed, v_k speed of block. (similarly for masses):

`m_t\cdot v_(t0)=m_t\cdot v_(t1) + m_k\cdot v_k\implies\\v_k=(m_t\cdot v_(t0)-m_t\cdot v_(t1))/(m_k)=(0.01kg\cdot(400-300)m/s)/(0.5kg)=2(m)/(s)`

The block will move with a speed of 2 m/s in the direction of the bullet.