You shoot a bullet with mass m1 and initial velocity v1 into a ballistic pendulu

Question

You shoot a bullet with mass m1 and initial velocity v1 into a ballistic pendulum. The wooden block of this pendulum has a mass of m2. The bullet undergoes a perfectly inelastic collision with the block and stays inside the block. The bullet-block object moves upward in a circular arc, allowed by 2 strings holding the block. The assembly rises to a maximum height h above the original height.

a. What is the initial momentum of the bullet?

b. What is the speed of the ball-block assembly immediately (infinitesimal time) after the bullet is embedded inside the block, before the assembly moves any appreciable distance?

c. What is the inital kinetic energy of the bullet-block assembly just before it starts to rise?

d. To what heigh h does the assembly rise?

I am a Physics noob, so I would very much appreciate it if you can try to not skip any steps to get to the solution.... Thanks in advance, and I'll try to post the picture if I can figure out how to.

Explanation / Answer

initial momentum=mass*velocity=m1*v1 b.using conservation of momentum principle, if speed of bullet+pendulum system is v, then m1*v1=(m1+m2)*v hence v=m1*v1/(m1+m2) c)just before it start to rise, initial kinetic energy=0.5*mass*speed^2=0.5*(m1+m2)*v^2 =0.5*(m1+m2)*m1^2*v1^2/(m1+m2)^2=0.5*m1^2*v1^2/(m1+m2) d.if it reaches height h, at maximum height it loses all its intital kinetic energy so all the kinetic energy gets converted to potential eenrgy= hence 0.5*m1^2*v1^2/(m1+m2)=(m1+m2)*g*h h=0.5*m1^2*v1^2/(g*(m1+m2)^2)

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