Where Worlds Collide... - 1D Collisions

HW Solutions:

handout questions:

1. Vf = 1 m/s, delta KE = -60,000 J

2. We can assume momentum is conserved at the moment of the collision, but no more, since gravity acts on the system of the ball and bullet.

final velocity of the bullet and ball is UPWARDS at 33.33 m/s.

Energy is conserved after the collision, with the initial KE = .5*(.150 kg + .030 kg)*(33.33 m/s)^2 and final U = mgh.

Solving, h = 56.69 m.

3. momentum is conserved:

(4 kg)(5 m/s) + (10 kg)(3 m/s) - (3 kg)(4 m/s) = (4 kg + 10 kg + 3 kg)*v_final

v_final = 2.235 m/s to the right.

From the textbook:

Questions:
8. The bat will be in contact with the ball for a longer time since, to hit it from home plate, it must have its velocity turned around. THis will make more sense after tomorrow's lesson.

11. Since the light body has much less gravity force, it must have a smaller mass than the heavy body. Since it has the same KE, this means its speed must be greater than that of the heavy body.

KE = (1/2)mv^2 = (1/2)*mv * v = (1/2)*p*v

Since the KE is the same for both, but the speed of the lighter object is greater, it means its momentum must go down. thus the more massive object will have a greater momentum, even though it is moving at a smaller speed.

12. If an object has no KE, it must have either zero speed or zero mass. In either case, the momentum must also be zero, so the answer is NO.

For the same reason, the opposite is false as well.

Problems:

32.

We know that the total energy is divided into two parts, by conservation of energy:

7500 J = .5*m*v^2 + .5*(1.5m)*v^2

From momentum, we can obtain that v2 = 2/3*v1.

Substituting, we can obtain that:

7500 J = .5*mv1^2 + (2/3)*.5*mv1^2

in other words, 7500 J = 1x + 2/3x where x = the energy of the smaller piece.

Solving, the smaller piece gets 4500 J, and the larger one gets 3000 J

69.

This problem is similar to #11 from yesterday. We can use newton's 2nd/work energy to find the initial speed of each car. We just have to do it twice.

We can find that Car A was going 13.28 m/s and Car B was going 18.78 m/s.

We know that car B was stationary before the crash, meaning it had no momentum.

m_a*v_a + 0 = m_a*v_{f_a} + m_b*v_{f_b}.

We can solve this for v_a = 22.7 m/s.

Since Car A skidded to this speed, we can again use work-energy to find the initial speed of the car:

-mu*m_a*g*delta_x = 1/2*m_a*(22.7 m/s)^2 - .5*m_a*vo^2

Solving for vo = 26.3 m/s. This, converted to miles per hour, is 58.83 MPH. The driver was speeding!