Question

A steel ball with mass m=5.21 g is moving horizontally with speed ????=412 m/s when it strikes a block of hardened steel with mass ????=14.8 kg (initially at rest). The ball bounces off the block in a perfectly elastic collision. (a) What is the speed of the block immediately after the collision? m/s ( ± 0.002 m/s) (b) What is the impulse exerted on the block?

Answer 1

Answer:

a) The speed of the block immediately after the collision is $v_{2f}=(0.289\±0.002)m/s$.

b) The impulse exerted on the block is $p_{2}=(4.2772\±0.0296)kg*m/s$.

Explanation:

Hi

a) As this is a perfectly elastic collision, we can use the formula  $v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}+ (\frac{m_{2}-m_{1}}{m_{1}+m_{2}} )v_{2i}$, due $v_{2i}=0m/s$, we obtain $v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}$. Then with the data that we know $m_{1}=5.21g=0.00521kg, m_{2}=14.8kg$ and $v_{1i}=412m/s$, therefore $v_{2f}=(\frac{2(0.00521kg) }{0.00521kg+14.8kg} ) 412m/s=0.289m/s$ or $v_{2f}=(0.289\±0.002)m/s$ adding uncertainty.

b) Now that we know the speed we can use $p_{2}=m_{2}*v_{f2} =14.8kg*(0.289\±0.002)m/s=(4.2772\±0.0296)kg*m/s.$