Question

A 2.5. Gram arrow enters a 100 g apple with a speed of 115 m/s. If the apple was originally at rest, then what speed will it have as the arrow enters it? Remember the formula is P=MxV

Answer 1

Answer:

The speed of the apple will be 2.81 m/s when the arrow enters it.

Explanation:                

We can find the speed of the apple by conservation of linear momentum:

$p_{i} = p_{f}$

$m_{ap}v_{ap_{i}} + m_{a}v_{a_{i}} = m_{ap}v_{ap_{f}} + m_{a}v_{a_{f}}$

Where:

$m_{ap}$ is the mass of the apple = 100 g = 0.1 kg

$m_{a}$ is the mass of the arrow = 2.5 g = 0.0025 kg

$v_{ap_{i}}$ and $v_{ap_{f}}$ is the initial and final speed of the  apple respectively

$v_{a_{i}}$ and $v_{a_{f}}$ is the initial and final speed of the  arrow respectively

Since the apple was originally at rest ($v_{ap_{i}}$ = 0) and knowing that $v_{a_{f}}$ = $v_{ap_{f}}$ when the arrow enters into the apple, we have:

$0 + 0.0025 kg*115 m/s = v(0.0025 kg + 0.1 kg)$  

$v = 2.81 m/s$  

Therefore, the speed of the apple will be 2.81 m/s when the arrow enters it.

I hope it helps you!