To solve this problem, we need to apply the principle of conservation of momentum. The total momentum before the jump is equal to the total momentum after the jump.

Before the jump, both the boy and the board are stationary, so the total momentum is 0.

After the jump, the boy is moving with a velocity of 15 m/s to the right. Let's denote the velocity of the board to the left as Vb.

The momentum of the boy is his mass times his velocity, or 50 kg * 15 m/s = 750 kg*m/s to the right.

The momentum of the board is its mass times its velocity, or 150 kg * Vb to the left.

Since the total momentum must be conserved, these two momenta must be equal in magnitude and opposite in direction. Therefore, we have:

750 kg*m/s = 150 kg * Vb

Solving for Vb gives Vb = 750 kg*m/s / 150 kg = 5 m/s.

Therefore, the boy and the board are separating from each other at a rate of 15 m/s (boy's speed) + 5 m/s (board's speed) = 20 m/s.

Question

To solve this problem, we need to apply the principle of conservation of momentum. The total momentum before the jump is equal to the total momentum after the jump.

Before the jump, both the boy and the board are stationary, so the total momentum is 0.

After the jump, the boy is moving with a velocity of 15 m/s to the right. Let's denote the velocity of the board to the left as Vb.

The momentum of the boy is his mass times his velocity, or 50 kg * 15 m/s = 750 kg*m/s to the right.

The momentum of the board is its mass times its velocity, or 150 kg * Vb to the left.

Since the total momentum must be conserved, these two momenta must be equal in magnitude and opposite in direction. Therefore, we have:

750 kg*m/s = 150 kg * Vb

Solving for Vb gives Vb = 750 kg*m/s / 150 kg = 5 m/s.

Therefore, the boy and the board are separating from each other at a rate of 15 m/s (boy's speed) + 5 m/s (board's speed) = 20 m/s.

Knowee AI · Accepted Answer

To solve this problem, we need to apply the principle of conservation of momentum. The total momentum before the jump is equal to the total momentum after the jump.

Before the jump, both the boy and the board are stationary, so the total momentum is 0.

After the jump, the boy is moving with a velocity of 15 m/s to the right. Let's denote the velocity of the board to the left as Vb.

The momentum of the boy is his mass times his velocity, or 50 kg * 15 m/s = 750 kg*m/s to the right.

The momentum of the board is its mass times its velocity, or 150 kg * Vb to the left.

Since the total momentum must be conserved, these two momenta must be equal in magnitude and opposite in direction. Therefore, we have:

750 kg*m/s = 150 kg * Vb

Solving for Vb gives Vb = 750 kg*m/s / 150 kg = 5 m/s.

Therefore, the boy and the board are separating from each other at a rate of 15 m/s (boy's speed) + 5 m/s (board's speed) = 20 m/s.

A boy of mass 50 kg stands on a board of mass 150 kg which in turn is kept on a frictionless horizontal ice surface. The boy makes a jump with a velocity component 15 m/s in horizontal direction with respect to the ice. With what rate in m/s are the boy and the board separating from each other?

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