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# Impulse Momentum, Conservation of Linear Momentum, & Collisions

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The problems came out from Physics 7th edition by Cutnell and Johnson, chapter 7. The sections include 1) the impulse-- momentum theorem 2) the principle of conservation of linear momentum 3) collisions in one dimension 4) collisions in two dimensions and 5) center of mass.

https://brainmass.com/physics/conservation-of-energy/impulse-momentum-conservation-linear-momentum-collisions-321756

#### Solution Preview

**Please consider significant figures and if possible, draw visual graphics**
Problems
(The Impulse Momentum)
 A golfer, driving a golf ball off the tee, gives the ball a velocity of +38 m/s. The mass of the ball is 0.045 kg, and duration of the impact with the golf club is 3.0 x 10-3 s. (a) what is the change in the momentum of the ball by the club.

Solution: Initial momentum of the ball = mvi = 0.045 x 0 = 0

Final momentum of the ball = mvf = 0.045 x 38 = 1.71 kg.m/s

 A volleyball is spiked so that its incoming velocity of +4.0 m/s is changed to an outgoing of -21 m/s. The mass of the volleyball is 0.35 kg. What impulse does the player apply to the ball?

Solution: Momentum of the incoming ball = mv = 0.35 x 4 = + 1.4 kg.m/s

Momentum of the outgoing ball = - 0.35 x 21 = - 7.35 kg.m/s

Impulse = Change in momentum = - 7.35 - 1.4 = - 8.75 kg.m/s
(The Principle of Conservation of Linear Momentum)

 A 5.5-kg swimmer is standing on a stationary 210 kg floating raft. The swimmer then runs off the raft horizontally with a velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.

Solution:: Initial momentum of the swimmer = Initial momentum of the raft = 0

Final momentum of the swimmer (at the instant he runs off the raft) = mv = 55 x 4.6 = 253 kg.m/s [mass of the swimmer as stated in the problem 5.5 kg which obviously is a typographical error. The mass has been assumed as 55 kg]

Let the recoil velocity of the raft be v. Then, final momentum of the raft (at the instant the swimmer runs off the raft) = 210v. By conservation of linear momentum we have:

Sum of the momentums of the swimmer and the raft at the instant the swimmer jumps off the raft = Sum of the momentums of the swimmer and the raft before the swimmer jumps off the raft

253 + 210v = 0 o v = - 1.2 m/s

Raft will move backwards at 1.2 m/s
(Collisions in One Dimension and Collisions in Two Dimensions)

 Kevin has a mass of 87 kg and skating with in-line skates. He sees his 22-kg younger brother up ahead standing on the sidewalk, with his back turned. Coming up from behind, he grabs his brother and rolls off at a speed of 2.4 m/s. Ignoring friction, find Kevin's speed just before he grabbed his brother.

Solution: Let initial speed of Kevin be v. Initial speed of brother is zero. Final speed of both together is 2.4 m/s.

Final momentum of Kevin & brother = Initial momentum of Kevin + Initial momentum of brother

(87 + 22) x ...

#### Solution Summary

Impulse momentum, conservation of linear momentum and collisions are examined in the solution.

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