A 44 kg child jumps off a 2.2 kg skateboard that was moving at 8.0 m/s. The skateboard comes to a stop as a result. Find the speed at which the child jumped from the board. Show all your work. Assume a frictionless, closed system.

Apply the law of conservation of momentum.

The child must have jumped on while traveling in the opposite direction
m*8.0 m/s + M*Vchild = (M + m) *Vfinal = 0

m is the skateboard mass and M is the child's mass.

Solve for Vchild

Vchild = -(m/M)*8.0 m/s

The child jumped off :)

whoops

i needz hlep

To find the speed at which the child jumped from the board, we can use the principle of conservation of momentum. According to this principle, the total momentum before the jump should be equal to the total momentum after the jump in a frictionless, closed system.

The momentum of an object is defined as the product of its mass and velocity. Mathematically, we can express momentum as:

momentum = mass × velocity

Before the jump, the 2.2 kg skateboard was moving at a velocity of 8.0 m/s. The momentum of the skateboard is:

momentum_skateboard_before = mass_skateboard × velocity_skateboard_before
= 2.2 kg × 8.0 m/s
= 17.6 kg·m/s

The child weighs 44 kg and jumps off the skateboard. Let's assume the child jumps with a velocity of v (which we need to find). The momentum of the child is:

momentum_child = mass_child × velocity_child
= 44 kg × v
= 44v kg·m/s

According to the principle of conservation of momentum, the total momentum before the jump should be equal to the total momentum after the jump, so we can set up the following equation:

momentum_skateboard_before + momentum_child = momentum_skateboard_after + momentum_child

17.6 kg·m/s + 44v kg·m/s = 0 kg·m/s + 44v kg·m/s

Simplifying the equation:

17.6 kg·m/s = 0 kg·m/s

This equation tells us that the momentum before the jump is equal to the momentum after the jump. Since the skateboard comes to a stop after the child jumps off, the momentum of the skateboard after the jump is 0 kg·m/s.

Thus, we can solve the equation for v:

17.6 kg·m/s = 0 kg·m/s + 44v kg·m/s
17.6 kg·m/s = 44v kg·m/s

Dividing both sides of the equation by 44 kg·m/s:

(17.6 kg·m/s) / 44 kg·m/s = (44v kg·m/s) / 44 kg·m/s
0.4 m/s = v

Therefore, the speed at which the child jumped from the board is 0.4 m/s.

original momentum = 46.2 * 8

Final momentum = original momentum
= 44 * v - 2.2 * 0
so
46.2 * 8 = 44 * v