Change of KE in Ballistic Pendulum Collision

[bobdream]

Homework Statement

Homework Equations

mv=(m+M)v'
KE_i=1/2mv²
PE=KE (after collision)

The Attempt at a Solution

I'm am particularly stuck on part C although I'm not very confident in my first two answers either.

a)
1/2(m+M)v_f²=(M+M)gh
v_f²=2gh
v_f=SQRT(2gh)

mv=(m+M)v_f
v=((m+M)SQRT(2gh))/m

b)
KE_i=m(m+M)²2gh)
-------------------
2m²

KE_f=1/2(M+m)2gh

deltaKE=1/2(M+m)2gh - m(m+M)²2gh)
-------------------
2m²

stuck...

Thanks for any help in advance.

 

[mighty2000]

Hello! For part C, you can use the conservation of momentum and energy equations to solve for the final velocity of the two objects after the collision. Since the objects are colliding and sticking together, the final velocity of both objects will be the same.

First, we can use the conservation of momentum equation: m1v1 + m2v2 = (m1 + m2)v' where v' is the final velocity of both objects after the collision. We can substitute in the values we know: m1 = m, m2 = M, v1 = ((m+M)*SQRT(2gh))/m, and v2 = 0 (since the second object is initially at rest). This gives us the following equation:

m*((m+M)*SQRT(2gh))/m + M*0 = (m+M)*v'

Simplifying, we get:

(m+M)*SQRT(2gh) = (m+M)*v'

Next, we can use the conservation of energy equation: KEi = KEf. We know the initial kinetic energy (KEi) is equal to the potential energy (PE) before the collision (since the objects are initially at rest). So, we can substitute in the values we know: KEi = 0, KEf = 1/2*(m+M)*v'^2, and PE = (m+M)*gh. This gives us the following equation:

0 = 1/2*(m+M)*v'^2 - (m+M)*gh

Simplifying, we get:

v'^2 = 2gh

Now, we can substitute this value for v'^2 into the first equation we found:

(m+M)*SQRT(2gh) = (m+M)*SQRT(2gh)

We can see that the two sides of the equation are equal, so this confirms that our value for v' is correct. Therefore, the final velocity of both objects after the collision is v' = SQRT(2gh).

I hope this helps! Let me know if you have any other questions. Good luck with your homework!