# Thread: Physics Problems: Conservation of Momentum

1. ## Physics Problems: Conservation of Momentum

I'm completely stumped on a few problems that are due after this weekend. If anyone could offer me hints on how to start them, it would be great. Thanks!

1. A 50 g bullet lodges into a 2.0 kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50 g mass just before entering?

2. A 0.150 kg bullet is fired at 715 m/s into a 2.0 kg wooden block at rest. The velocity of the block afterward is 40 m/s. The bullet passes through the block and emerges with what velocity?

3. A 2.0 kg ball moving to the right at 1.0 m/s strikes a 4.0 kg ball moving left at 3.0 m/s. What are the velocities after impact, assuming complete elasticity?

4. A 2.0 kg block A and a 1.0 kg block B are pushed together against a spring and tied with a cord. When the cord breaks, the 1.0 kg block moves to the right at 8.0 m/s. What is the velocity of the 2.0 kg block?

Edit: It would appear the OP now only requires help with Questions 1. and 3.

2. Originally Posted by xxlvh
I'm completely stumped on a few problems that are due after this weekend. If anyone could offer me hints on how to start them, it would be great. Thanks!

1. A 50 g bullet lodges into a 2.0 kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50 g mass just before entering?

2. A 0.150 kg bullet is fired at 715 m/s into a 2.0 kg wooden block at rest. The velocity of the block afterward is 40 m/s. The bullet passes through the block and emerges with what velocity?

3. A 2.0 kg ball moving to the right at 1.0 m/s strikes a 4.0 kg ball moving left at 3.0 m/s. What are the velocities after impact, assuming complete elasticity?

4. A 2.0 kg block A and a 1.0 kg block B are pushed together against a spring and tied with a cord. When the cord breaks, the 1.0 kg block moves to the right at 8.0 m/s. What is the velocity of the 2.0 kg block?
Some random thoughts to be applied (after due thought) where appropriate:

Think about conservation of mechanical energy after impact and think about conservation of momentum during impact.

Elastic collision means kinetic energy of the system is conserved.

3. So for the elastic collision the total sum of the momentum and kinetic energy must be equal after the equation. I don't know how to set up the equation to solve though, since none of the final velocities were given..

4. Originally Posted by xxlvh
[snip]3. A 2.0 kg ball moving to the right at 1.0 m/s strikes a 4.0 kg ball moving left at 3.0 m/s. What are the velocities after impact, assuming complete elasticity? [snip]
Originally Posted by xxlvh
So for the elastic collision the total sum of the momentum and kinetic energy must be equal after the equation. I don't know how to set up the equation to solve though, since none of the final velocities were given..
I don't see the trouble here. The final velocities are the things you're trying to find. So you give them a symbol like u and v and substitute into the appropriate equations.

$\displaystyle 2(1) + 4(-3) = 2u + 4v$ .... (A)

$\displaystyle \frac{1}{2} (2) (1)^2 + \frac{1}{2} (4) (3)^2 = \frac{1}{2} (2) u^2 + \frac{1}{2} (4) v^2$ .... (B)

I'll let you figure which equation corresponds to which conserved quantity, and you should obviously simplify the equations. Then solve equations (A) and (B) simultaneously.

5. Thank you, I'll work on finishing that question up then... any suggestions for problem #1?

6. 1. A 50 g bullet lodges into a 2.0 kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50 g mass just before entering?

Energy should be conserved - I hope. So:
Kinetic E of bullet = Potential E of {bullet+block}
1/2 * m_b * V^2 = (m_b+m_Block)*g*height

Find V above.

Be careful about the conversion of units... g, kg, cm, m, etc...
and unit of g=m/s^2