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__Impulse__

1) A gold ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5 x 10^-3 s. Find (a) the impulse imparted to the golf ball and (b) The average force exerted on the ball by the golf club.

**Answer: (a) 2.0 kg(m/s) (b) 15.8 x 10^2 N**

My approach....

(a) Impulse = velocity x change in mass = (0.045 kg)(45 m/s) = 2.0 kg(m/s) Got the first part.

(b) To get the average force would I not take impulse divided by time? 2.0/3.5 x 10^-3 = 571... but the answer is 15.8. x 10^2

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__Elastic Collisions__

2) A 0.450 kg ice puck, moving east with a speed of 3.0 m/s has a head on collision with a 0.900 kg puck intially at rest. Assuming a perfectly elastic collision what will be the speed and direction of each object after the collision?

**Answer: 1.00 m/s W & 2.00 m/s E**

My thoughts were to just use the momentume formula...

(.5)(.450kg)(3 m/s) + (.5)(.900 kg)(0) = (.5)(.450kg)(v) + (.5)(.900 kg)(v)

But how do I determine it with two unknowns?

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__Inelastic Collisions__

3) A 920 kg sports car collides into the rear end of a 2300 kg SUV stopped at a red light. The bumber's lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, knowing that the coeffiecent of kinetic friction b/t tires and road is 0.80 calculates the speed of the sports car at impact. What was that speed?

**Answer: 23 m/s**

Again I was going to use the same approach... but I had no speeds. Please someone help...