Momentum and Conservation

As promised, here is the solution to the first problem on the sheet. Since I said to the right was positive, any momentum going to the right is positive.

Note the importance of setting the final momentum of the bowling ball as NEGATIVE in part (b) since it is going to the left.



2. Chippy and Skippy:
Since the ice is frictionless, momentum is conserved. Let's set to the right as positive, and assume that Skippy goes to the right.

P_o = P_f
0 + 0 = (2 kg)(1.5 m/s) + (.4 kg)*v_f

Solving, v_f = -7.5 m/s. The negative sign means that Chippy is going to the left.

3. We can again assume momentum is conserved here, as otherwise, the problem could not be solved.
P_o = P_f
m*v + 0 = (m)(-.5v) + (5m)*v_f

v_f = 0.3v

4.
m_bullet*v_o + 0 = (m_bullet + m_block)*v_f

Since the bullet is stuck in the block, they both go at the same velocity v_f = 0.5 m/s.

Solving for the initial speed of the bullet, v_o = 167.2 m/s

HW solutions:
Question 4. If the rich man threw the bag of gold coins in one direction, since momentum is conserved on the frictionless ice, he would slide in the opposite direction, eventually getting to shore.

The loophole to this story is that the gold coins would eventually reach the shore of the lake. Once the rich man also arrived at shore, he could run around the border of the lake to grab his bag of coins. This wouldn't make for as good of a story, however.

Problems:
3. Skip this question until tomorrow - we had to skim past this in class due to time constraints.

4. We can say that momentum is conserved IMMEDIATELY after the collision.

(95 kg)(4.1 m/s) + (85 kg)(5.5 m/s) = (95 kg + 85 kg)*v_final

v_final = 4.76 m/s

5. Momentum is conserved in the horizontal direction.

(12,500 kg)(18.0 m/s) = (12,500 kg + 5750 kg)*v_final

v_final = 12.33 m/s

6. Momentum is conserved.
(9500 kg)*(16 m/s) + 0 = (9500 kg)*6.0 m/s + m*(6.0 m/s)

m = 15,833 kg

8. This problem can be solved in two pieces. Since the block skidded to a stop, we can use either work-energy principle or Newton's 2nd law to figure out the initial speed of the block.

Using either method, we can obtain that v_o = (2*g*mu*delta_x)^(1/2) = 6.823 m/s.

During the collision with the block, momentum is conserved. We now let the initial speed of the block and bullet become the FINAL speed of the collision between the bullet and the block. (v_f = 6.823 m/s)

m_bullet*v_o + 0 = (m_bullet + m_block)*v_f

v_o = (m_bullet + m_block)*v_f/(m_bullet) = 507.2 m/s

11. Momentum is conserved.

(.013 kg)*(230 m/s) + 0 = (2.0 kg)*v_f + (.1023 kg)*(170 m/s)

v_f = 0.39 m/s

Problems 16 and 17 will make more sense later on - don't worry about them for now.