Question

1. Suppose rain falls vertically into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumlating water, the kinetic energy of the cart.

a. increases
b. does not change
c. decrease

2. two marbles, one twice as heavy as the other, are dropped to the ground from the roof of a building. Just before hitting the ground, the heavier marble has

a. as much kinetic energy as the lighter one
b. twice as much kinetic energy as the lighter one
c. half as much kinetic energy as the lighter one
d. four times as much kinetic energy as the lighter one
e. impossible to determine

3. A person attempts to knock down a large wooden bowling pin by throwing a ball at it. The person has two balls of equal size and mass, one made of rubber and the other of putty. The rubber ball bounces back, while the ball of putty sticks to the pin. Which ball is most likely to topple the bowling pin?

a. the rubber ball
b. the ball of putty
c. makes no difference

4. We launch a 1 kg block with spring along a rough surface (coefficient of kinetic friction of 0.45).
a.If a spring compression of 0.25m launches the block a total of 2.5m, what is the spring constant of the spring?
b.How much compression is required to reach a target 5.0m away?

Answer 1

1. C. There are no external horizontal forces, so the horizontal momentum of the cart before the interaction is the same as the momentum of the cart and water after the interaction. The momentum is unchanged but the mass increases. Using KE = p^2/(2m), the kinetic energy decreases.

2. b. twice as much kinetic energy as the lighter one

3. a. the rubber ball

4. First, let's assume that the surface is horizontal, so friction = u*m*g.

The spring energy that is transferred into the block is absorbed by the work of friction. Relate those to find the spring constant.
1/2*k*x^2 = friction*d = u*m*g*d

The block was launch a total of 2.5m. This isn't very well worded, because they don't really tell you where the starting point is at. Is it measured from the full compression location or from 0 position? I'm just going to assume they mean it starts from the full compression location, so d = 2.5.

Solve for k
k = 2*u*m*g*d / x^2

Plug in numbers
I got, k = 353 N/m^2

Just use the same relation, solved for x
x^2 = 2*u*m*g*d/k
x = sqrt(of all that)

Plug in numbers
I got, x = 0.354 m