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Homework Help: Some Study Questions For PHYS100A

  1. Mar 17, 2006 #1
    Here are some study questions for a college Physics 100A course I dug up. They seem to be related to momentum. Let's see how well you can do:
    A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 56.0 m/s and returns the shot with the ball traveling horizontally at 42.0 m/s in the opposite direction. (Assume the initial direction of the ball is in the -x direction.)

    (a) What is the impulse delivered to the ball by the racket?
    (b) What work does the racket do on the ball?
    A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 15 m/s. The ball is hit straight back at the pitcher with a final speed of 28 m/s. (Assume the direction of the initial motion of the baseball to be positive.)
    (a) What is the impulse delivered to the ball?
    (b) Find the average force exerted by the bat on the ball if the two are in contact for 2.0 10-3 s.
    A 750 N man stands in the middle of a frozen pond of radius 7.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2 kg physics textbook horizontally toward the north shore at a speed of 7.0 m/s. How long does it take him to reach the south shore?
    A 7.0 g bullet is fired into a 1.5 kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 9.3 cm. Find the initial speed of the bullet.
    A 28.0 g object moving to the right at 19.0 cm/s overtakes and collides elastically with a 6.0 g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision.
    28.0 g object cm/s
    6.0 g object cm/s
    A billiard ball rolling across a table at 1.40 m/s makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision when each of the following occurs.
    (a) The second ball is initially at rest.
    1st ball m/s
    2nd ball m/s
    (b) The second ball is moving toward the first at a speed of 1.10 m/s.
    1st ball m/s
    2nd ball m/s
    (c) The second ball is moving away from the first at a speed of 0.95 m/s.
    1st ball m/s
    2nd ball m/s
    A 89 kg fullback moving east with a speed of 8.0 m/s is tackled by a 98 kg opponent running north at 2.0 m/s. If the collision is perfectly inelastic, calculate each of the following.
    (a) the velocity of the players just after the tackle
    (b) the kinetic energy lost as a result of the collision
    Can you account for the missing energy?
    A billiard ball moving at 5.30 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.54 m/s, at an angle of 31.0° with respect to the original line of motion.
    (a) Find the velocity (magnitude and direction) of the second ball after collision.
    ° (with respect to the original line of motion, include the sign of your answer; consider the sign of the first ball's angle)
    (b) Was the collision inelastic or elastic?

    A 11.2 g bullet is fired horizontally into a 105 g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact?
    A m1 = 0.400 kg block is released from rest at the top of a frictionless track h1 = 2.10 m above the top of a table. It then collides elastically with a 1.00 kg block that is initially at rest on the table
    (a) Determine the velocities of the two objects just after the collision.
    velocity of m1 m/s
    velocity of m2 m/s
    (b) How high up the track does the 0.400 kg object travel back after the collision?
    (c) How far away from the bottom of the table does the 1.00 kg object land, given that the table is 2.10 m high?
    (d) How far away from the bottom of the table does the 0.400 kg object eventually land?
  2. jcsd
  3. Mar 17, 2006 #2


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    Gold Member

    Sounds like homework....
  4. Mar 17, 2006 #3
    First question I'll only help with.. It seems like too much for you to be stuck on :-|

    If we have a velocity in <- that direction and it bounces of in -> direction.

    Assume -> direction is positive and <- is negative.

    dV (delta Velocity, change in velocity) is Vf-Vi which is +V - (-V) which is V+V.

    Don't just simply subtract, velocity is a vector.
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