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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.

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    SOLUTION This solution is FREE courtesy of BrainMass!

    **Please consider significant figures and if possible, draw visual graphics**
    (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 2.4 = 87v + 0 or v = 3 m/s

     A 1055-kg van stopped at a traffic light, is hit directly in the rear by a 715 -kg car traveling with a velocity of +2.25 m/s Assume that the transmission of the van is neutral, the brakes are not being applied and the collision is elastic. What is the final velocity of (a) the car and (b) the van?

    Solution: Initial momentum of the van = 0 (as it is at rest)

    Initial momentum of the car = 715 x 2.25 = 1609 kg.m/s

    Let the final velocities of the van and the car be vv and vc.

    By conservation of linear momentum: 1055vv + 715vc = 1609

    1.48vv + vc = 2.25 ........ (1)

    As the collision is elastic, the kinetic energy of the two objects is also conserved. Hence,

    KE of van after collision + KE of car after collision = KE of van before collision + KE of car before collision

    ½ x 1055 x (vv)2 + ½ x 715 x (vc)2 = ½ x 715 x 2.252 + 0

    1055 x (vv)2 + 715 x (vc)2 = 3620

    1.48(vv)2 + (vc)2 = 5.06 ........(2)

    From (1) (vc)2 = (2.25 - 1.48vv)2 .........(3)

    From (2) (vc)2 = 5.06 - 1.48(vv)2 ..........(4)

    Equating (3) and (4): (2.25 - 1.48vv)2 = 5.06 - 1.48(vv)2

    5.06 + 2.2(vv)2 - 6.66vv = 5.06 - 1.48(vv)2

    3.68(vv)2 = 6.66vv

    vv = 6.66/3.68 = 1.8 m/s

    Substituting in (1): 1.48 x 1.8 + vc = 2.25

    vc = - 0.41 m/s

    a) Car will recoil at a speed of 0.41 m/s

    b) Van will move forward with a speed of 1.8 m/s

     A 2.50 gram bullet, traveling at a speed of 425 m/s strikes the wooden block of a ballistic pendulum, such as the figure below. The block has a mass of 2.15 gram. (a) Find the speed of the bullet/block combination immediately after the collision. (b) How high does the combination rise above its initial position?
    The Physics of measuring the speed of a bullet

    (a) A bullet approaches a ballistic pendulum



    (b) The block and the bullet swing upward after the collision
    Hf = 0.650 m

    m1 + m2



    Initial momentum of the block before collision = 0

    Initial momentum of the bullet before collision = v01m1

    Final momentum of block + bullet = (m1+m2)vf

    By conservation of linear momentum: Net momentum before collision = Net momentum after collision

    v01m1 = (m1+m2)vf

    Or vf = v01m1/(m1+m2)

    As the block with bullet swings up, its KE is converted into its PE. Let the final height reached by the block be Hf. Then,

    KE at the lowest point of the block = PE at the highest point

    ½ (m1+m2) [v01m1/(m1+m2)]2 = (m1+m2)gHf

    ½ [v01m1/(m1+m2)]2 = gHf

    Hf = (1/2g)[v01m1/(m1+m2)]2

    Substituting correct values Hf can be determined.

     A car (mass = 1100 kg) is traveling at 32 m/s when it collides head-on with a sport utility vehicle (mass = 2500 kg) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the utility vehicle traveling?

    Solution: Initial momentum of car = 1100 x 32 = 35200 kg.m/s

    Let speed of the sports vehicle = v. Then, initial momentum of sports vehicle = - 2500v (-ve sign indicates sports vehicle traveling in opposite direction).

    After collision momentum of two vehicles = 0

    By conservation of momentum: Net initial momentum = Net final momentum

    35200 - 2500v = 0 or v = 14.08 m/s

     A 5.00-kg ball, moving to the right at a velocity of +2.00 m/s on a frictionless table, collides heads-on with a stationary 7.50 -kg ball. Find the final velocities of the balls if the collision is (a) elastic and (b) completely inelastic.

    2 m/s
    5 kg 7.5 kg
    Ball 1 Ball 2

    a) Elastic collision

    Let the velocities of the two balls after collision be v1 and v2.

    Initial momentum of ball 1 = 5 x 2 = 10 kg.m/s

    Initial momentum of ball 2 = 0

    Final momentum of ball 1 = 5v1

    Final momentum of ball 2 = 7.5v2

    By conservation of momentum: Net momentum after collision = Net momentum before collision: 5v1 + 7.5v2 = 10

    v1 + 1.5v2 = 2 .......(1)

    By conservation of KE: Net KE after collision = Net KE before collision

    ½ x 5 x v12 + ½ x 7.5 x v22 = ½ x 5 x 22

    5v12 + 7.5v22 = 20

    v12 + 1.5v22 = 4 ........(2)

    From (1) v12 = (2 - 1.5v2)2 ......(3)

    From (2) v12 = 4 - 1.5v22 ......(4)

    Equating (3) and (4): (2 - 1.5v2)2 = 4 - 1.5v22

    4 + 2.25v22 - 6v2 = 4 - 1.5v22

    3.75v22 = 6v2

    v2 = 1.6 m/s

    Substituting in (1): v1 = 2 - 1.5 x 1.6 = - 0.4 m/s

    b) Completely inelastic collision

    Two balls join together to form one object after collision. Let velocity of the two balls together be v. Then, from conservation of momentum:

    Net momentum after collision = Net momentum before collision: (5 + 7.5)v = 10

    v = 10/12.5 = 0.8 m/s
    Conceptual Questions:
     Two identical automobiles have the same speed, one traveling east and one traveling west. Do these cars have the same momentum? Explain.

    Answer: No. Momentum is a vector quantity. Their momentums have same magnitude but their directions are opposite.

     Two objects have the momentum. Do the velocities of these objects necessarily have (a) the same directions and (b) the same magnitudes? Give your reasoning at each case.

    Answer: Two vectors are said to be equal if their magnitudes are same and their directions are same. Momentum being a vector, two objects will have same momentum only if a) their velocities have the same direction and b) product of the mass and velocity of each is same (i.e. the velocities need not be same but product of the mass and velocity of each object must be same).

     a) Can a single object have kinetic energy but no momentum? (b) Can a single object have a kinetic energy nut no momentum? (b) Can a system of two or more objects have a total kinetic energy that is not zero but a total momentum that is zero? Account for your answer.

    Answer: a) No. A single object must have a velocity to have kinetic energy and in that case it will have momentum also. b) Yes that is possible. If two objects have same magnitude of momentum but opposite directions, their combined momentum of the system of two objects will be zero. However, each will have a certain KE, sum of which will give a non zero KE.

     A satellite explodes in outer space, far from any other body sending thousands of pieces in all directions. How does the linear momentum of the satellite before the explosion compare with the total linear momentum of all the pieces after explosion? Account for your answer.

    Answer: For an isolated system (i.e. a system of objects not subjected to any external force) net linear momentum of the objects comprising the system is conserved (i.e. remains constant). The satellite being far away from any other body constitutes an isolated system (explosion takes place due to internal forces). Hence, sum of the linear momentums of the pieces will be same as the linear momentum of the satellite before explosion.

     In an elastic collision, is the kinetic energy of each object the same before and after the collision? Explain.
    Answer: In an elastic collision, kinetic energy of each object need not be same before and after collision. However, the sum of the kinetic energies of different objects will before and after collision will be same.

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