Question

A bomb initially at rest explodes into three pieces. The first piece is 0.25 kg and moves

to the right at 65 m/s. The second piece is 0.5 kg and moves to the right at 30 m/s. If the
third piece has a mass of 1.0 kg, what is its velocity after the bomb explodes?

Answer 1

The velocity of the third piece after the bomb explodes is -31.25 m/s to the left.

Step-by-step explanation:

To determine the velocity and direction of the third piece after the bomb explodes, we need to apply the law of conservation of momentum. According to this law, the total momentum before the explosion is equal to the total momentum after the explosion.

Let's consider the momentum of each piece before the explosion:

Piece 1:

momentum = mass × velocity = 0.25 kg × 65 m/s = 16.25 kg×m/s

Piece 2:

momentum = mass × velocity = 0.5 kg × 30 m/s = 15 kg×m/s

To find the momentum of the third piece, we subtract the total momentum of the first two pieces from the momentum of the bomb before the explosion:

Momentum of the bomb = 0 kg×m/s (since the bomb is at rest)

Total momentum of the first two pieces = 16.25 kg×m/s + 15 kg×m/s = 31.25 kg×m/s

Momentum of the third piece = Momentum of the bomb - Total momentum of the first two pieces

Momentum of the third piece = 0 kg×m/s - 31.25 kg×m/s = -31.25 kg×m/s

Since momentum is a vector quantity, the negative sign indicates that the third piece moves in the opposite direction of the first two pieces. Therefore, the velocity of the third piece after the bomb explodes is -31.25 m/s to the left.