Answer to Question #290397 in Mechanics | Relativity for Lebohang

Question #290397

A 10 g bullet travels horizontally at 2000 m•s-¹ passung through a 2kg pendulum which then moves initially horizontally at 4m•s-¹. Assume that the pendulum loses no mass as the bullet moves through it. Calculate the final speed of the bullet.

1
Expert's answer
2022-01-25T07:13:38-0500

Gives

m1=0.01kgv1=2000m/secm2=2kgv2=4m/secm_1=0.01kg\\v_1=2000m/sec\\m_2=2kg\\v_2=4m/sec

We know that

m1v1+m2v2=(m1+m2)Vm_1v_1+m_2v_2=(m_1+m_2)V


V=0.01×2000+2×40.01+2=282.01=13.93m/secV=\frac{{0.01\times2000}+2\times4}{{0.01+2}}=\frac{28}{2.01}=13.93m/sec


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment