Consider an elastic collision between two objects of equal mass, with one of the objects at rest initially (think about hitting a pool ball with the cue ball). Show that, after the collision, the velocities between the two equal-mass objects will always be perpendicular to each other. Hint: Use conservation of momentum to get an equation for the vector sum of the two final velocities, and square this equation (i.e. take the dot product of each side with itself). Then use conservation of kinetic energy and compare your equations.

A particle with a mass of 1.20 kg is acted on by a force Fx acting in the x-direction. If the magnitude of the force varies in time as shown in the figure below, determine the following. (a) Impulse of the force (in kg · m/s ) (b) Final velocity of the particle (in m/s) if it is initially at rest (c) Find the final velocity of the particle (in m/s) if it is initially moving along the x-axis with a velocity of −2.30 m/s.

The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were not present. For example, consider a 58 kg person traveling at 26 m/s (about 58 mph). Use impulse and momentum to calculate how many newtons of force would be required to stop the person in 0.039 s, 0.078 s, and 0.39 s. Note how the force changes with increased time. (Enter the magnitudes.)   (a) How much force is needed to stop the person in 0.039 seconds?N (b) How much force is needed to stop the person in 0.078 seconds? N (c) How much force is needed to stop the person in 0.39 seconds? N

A particle with a mass of 1.40 kg is acted on by a force Fx acting in the x-direction. If the magnitude of the force varies in time as shown in the figure below, determine the following.   (a) impulse of the force (in kg · m/s ) (b) final velocity of the particle (in m/s) if it is initially at rest (c) Find the final velocity of the particle (in m/s) if it is initially moving along the x-axis with a velocity of  −2.70 m/s.

A 190-gram softball pitched at 22 m/s (horizontally) batted on a horizontal line drive back to the pitcher at 34 m/s. a.) How large is the magnitude of the impulse imparted to the ball by the bat? b.)The bat contacts the ball during a time of 5.6 ms. What magnitude average force did the ball experience?

the right. They undergo a head-on elastic collision and each returns, retracing its original path. Find the final speeds of the particles. (Enter your answers in terms of v;, the initial speed of each mass.) Two blocks with masses 6m and 9m are on a collision course with the same initial speeds v;. The block with mass 6m is traveling to the left, and the 9m block is traveling to V6m,f - V; V9m,f V; %3D Need Help? Read It

The x-component of a force on a 15-g golf ball by a golf club versus time is plotted in the attached figure.   a. Find the x-component of the impulse, in newton-seconds, during the following interval: [50 msec, 100 msec]  b. Find the change in the x-component of the momentum, in kilogram meters per second, during the following interval: [50 msec, 100 msec]

For the same collision as above, motion sensors record the velocities as a function of time for both carts. The graph of this function for the heavier cart is this: 2 t What qualities can we expect to see for a graph of velocity vs. time coming from the motion sensor recording the lighter cart's motion during the same collision? [Note: In our lab, the motion sensors are looking at their respective carts from opposite directions, so the positive direction for one sensor is the negative direction for the other, but for the purposes of this question, we will assume that the sensors agree upon the positive direction .] O The graph goes no higher than dotted line #2. The graph crosses dotted line #2, but goes no higher than dotted line #1. The graph crosses both dotted lines. There is not enough information to determine whether the graph crosses either dotted line.

Based on the principle conservation of linear momentum, which statement is true ? (Ma and Mp are masses of two objects a and b. Vaj and Vb; are the initial velocities of Mą and Mp respectively before collision. Vaf and Vbf are the final velocities of Ma and Mp respectively after collision.) MaVbit MoVai =MaVbpt MoVas None of the options available MVai + MBVbi = MVaf + MpVbf MaVai+MpVbf=MaVaf+ MbVbi MaVai+ MaVas=MBVbi+ MbVbf

Now consider the interactions between the two objects that undergo a collision. Apply the Impulse-Momentum Theorem to explain the changes in the momentum of the individual objects and that of the entire system. Use appropriate physics terminology in your discussion. As a matter of semantics, we say that a force is exerted on an object by another object; an impulse is imparted to an object by another object; an object or system can have, gain, or lose momentum; momentum is transferred from one object or system to another object or system.