1) A 3.04 kg particle has a velocity of (2.90 i - 4.04 j) m/s. Find the magnitude and direction of its momentum (counterclockwise from the positive x axis).

I found the magnitude, but for some reason I'm having problems with the direction. I found the angle to be 54.3 degrees by using cosine and then added 180 degrees to it because it is in the third quadrant.

2) Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig. P9.6). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. Find the original elastic energy in the spring if M = 0.300 kg.

This one I am officially confused on. Don't you need to use (1/2)kx^2 to get the energy of a spring? What does mass have to do with this? An earlier part to this problem asked for the speed of the block of mass M, which I found to be -6 m/s. I don't know if that has anything to do with this.

3) A garden hose full of motionless water is held in the manner shown in Figure P9.13. What additional force is necessary to hold the nozzle stationary if the discharge rate is 0.592 kg/s with a speed of 23.4 m/s?

I multiplied the change in momentum by the velocity and got the correct answer (13.9 N), but I'm not quite sure why it worked. Isn't linear momentum mass times velocity and impulse force times the change in time? I didn't have a change in time and the two formulas didn't seem to go together with the information provided.

4) A 7.00 kg bowling ball collides head-on with a 2.00 kg bowling pin. The pin flies forward with a speed of 3.01 m/s. If the ball continues forward with a speed of 1.77 m/s, what was the initial speed of the ball? Ignore rotation of the ball.

Here I thought I could use the formula for an elastic collision. I got

(1/2)(7)v^2 + (1/2)(2)(0) = (1/2)(7)(1.77^2) + (1/2)(2)(3.01^2)

and got 2.39 m/s for the velocity, but I guess that isn't right.

5) Consider a frictionless track ABC with a height of 5.00 m (A) going down to a height of 0 m (B). A block of mass m1 = 4.98 kg is released from A. It makes a head-on elastic collision at B with a block of mass m2 = 9.10 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

I got the velocity to be 33.3 m/s, but I don't know if that helps at all. I wasn't quite sure how to attack this problem because I didn't understand why m1 would reach a different height at all .. would it be zero??

6) A 89.5 kg fullback running east with a speed of 5.06 m/s is tackled by a 95.7 kg opponent running north with a speed of 3.08 m/s. If the collision is perfectly inelastic, calculate the speed and direction of the players just after the tackle.

I got the speed to be 2.92 m/s. I'm just having problems with the angle again. I got, using sine, the angle to be 33.0 degrees, so I subtracted that from 90 and then added 90 to it considering the direction the magnitude was facing and got 147 degrees.

7) During the battle of Gettysburg, the gunfire was so intense that several bullets collided in mid-air and fused together. Assume a 4.90 g Union musket ball was moving to the right at a speed of 244 m/s, 19.6° above the horizontal, and that a 3.07 g Confederate ball was moving to the left at a speed of 279 m/s, 14.1° above the horizontal. Immediately after they fuse together, what is their velocity?

I thought I could use vf = (m1v1i + m2v2i) / m1+m2, and got vf to be 246 m/s by incorporating the cosine of the angles into their corresponding velocities. The problem is, I need a velocity on the x axis and on the y axis and I was confused as to how to split them up.

8) Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in a perfectly elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.20 m/s. After the collision, the orange disk moves along a direction that makes an angle of 39.0° with its initial direction of motion and the velocity of the yellow disk is perpendicular to that of the orange disk (after the collision). Determine the final speed of each disk.

This one plain confuses the hell out of me.