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Physics

1. A 4.0-kg block slides along a frictionless surface with a constant speed of 5.0 m/s. Two seconds after it begins sliding, a horizontal, time-dependent force is applied to the mass. The force is removed eight seconds later. The graph shows how the force on the block varies with time.
A. What is the magnitude of the total impulse of the force acting on the block?

B. What approximately, is the speed of the block at t=11 seconds

2. A bullet of mass m is fired at speed v0 into a wooden block of mass M. The bullet instantaneously comes to rest in the block. The block with the embedded bullet slides along a horizontal surface with a coefficient of kinetic friction u.
Which one of the following expressions determines how far the block slides before it comes to rest?

3. In the game of billiards, all the balls have approximately the same mass, about 0.17 kg. In the figure, the cue ball strikes another ball such that if follows the path shown. The other ball has a speed of 1.5 m/s immediately after the collision. What is the speed of the cue ball after the collision?

pur-new-sol

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Please see the attached file.

A. What is the magnitude of the total impulse of the force acting on the block?

B. What approximately, is the speed of the block at t=11 seconds
Af the end of 10th second, the velocity is

Then the block will continue to slide on the frictionless surface. So the speed at 11 seconds is 15.5 m/s as well.

Next Question:

First let's take a look at the collision process. According to conservation of momentum for the very short duration impact,
, where vi is the speed of the block and the bullet just after the collision.
Then with this speed, the combo slides a distance of s to stop. In this process, according to the work-energy theorem, the work done by the frictional force is equal to the change of kinetic energy.
The work done is
The change of kinetic energy is

So

So c is correct.

Next Question on page 2!!

According to the conservation of momentum, in the y-direction

So the speed of the cue ball is 2.6m/s.
You can also solve it by taking the x-direction momentum.