Question

an arrow of mass m is shot horizontally at speed v. the arrow passes through the center of mass of melon with mass M. calculate the speed of the arrow-melon system just after the arrow sticks in the melon

Answer 1

The final velocity of the melon-arrow system immediately after collision is:

`v_f=(m*v)/((m+M))`

Step-by-step explanation:

We use conservation of momentum to solve this problem.

The initial state consists of an arrow of mass m and speed v , and a static melon that is not moving (velocity = 0)

Therefore, the initial momentum
`P_i` of the system which is the addition of the initial momentum of the arrow (
`p_(ai)`) plus the initial momentum of the melon (
`p_(mi)` is;

`P_i = p_(ai)+p_(mi)\\P_i=m*v+M*0\\P_i=m*v`

The final system consists of the arrow stack to the melon (total mass "m+M"), travelling at the unknown velocity
`v_f` that we need to find. The final momentum of this system is therefore the product of this mass times the unknown velocity:

`P_f=(m+M)*v_f`

Due to conservation of momentum in this inelastic collision, we set the equation that equals the system's initial momentum to the final momentum, and solve for the unknown velocity:

`P_i=P_f\\m*v=(m+M)*v_f\\v_f=(m*v)/((m+M))`