Question

A boy standing on a skateboard throws a 3.00kg gift box horizontally with a velocity of 14.5 m/s. The skateboard was initially at rest. Calculate the velocity of the boy and skateboard immediately after the box is thrown. The total mass of the boy and skateboard is 46.0kg. Assume rightward is positive.

Answer 1

The velocity of the boy and skateboard immediately after the box is thrown is -0.364m/s to the left.

Step-by-step explanation:

To find the velocity of the boy and skateboard immediately after the box is thrown, we can use the principle of conservation of momentum.

Initially, the total momentum of the system (boy, skateboard, and box) is zero since everything is at rest. After the box is thrown, the total momentum is still zero since there is no external force acting on the system.

Since momentum is conserved, the momentum of the box in one direction is equal to the momentum of the boy and skateboard in the opposite direction. Using the equation:

m1v1 + m2v2 = -m3v3

where m1, v1 represents the mass and velocity of the boy and skateboard, m2, v2 represents the mass and velocity of the box, and m3, v3 represents the mass and velocity of the boy and skateboard after the box is thrown.

Substituting the given values into the equation:

(46.0kg)(0m/s) + (3.00kg)(14.5m/s) = (-46.0kg - 3.00kg)v3

After solving for v3, we find that v3 is -0.364m/s.

Since rightward is defined as positive, the velocity of the boy and skateboard immediately after the box is thrown is -0.364m/s to the left.