Problem #1
A 3.00 kg particle has a velocity of (3.0i − 4.0j) m/s. Find its x and y components of momentum and the magnitude of its total momentum. |
| Fig.1 |
Answer:
Using the definition of momentum and the given values of m and v we have:
p = mv = (3.00 kg)(3.0i − 4.0j) m/s = (9.0i − 12.j) kg·m/s
So the particle has momentum components
px = +9.0 kg·m/s and py = −12. kg·m/s .
The magnitude of its momentum is
p = [px2 + py2]1/2 = [(9.0)2 + (−12.)2]1/2 kg·m/s = 15. kg·m/s
Problem #2
A 3.0kg steel ball strikes a wall with a speed of 10 m/s at an angle of 600 with the surface. It bounces off with the same speed and angle, as shown in Fig. 1. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the wall by the ball?
Answer:
The average force is defined as F = Δp/Δt, so first find the change in momentum of the ball. Since the ball has the same speed before and after bouncing from the wall, it is clear that its x velocity (see the coordinate system in Fig.1) stays the same and so the x momentum stays the same. But the y momentum does change. The initial y velocity is
viy = −(10 m/s ) sin 600 = −8.7 m/s
and the final y velocity is
vfy = +(10 m/s ) sin 600 = +8.7 m/s
so the change in y momentum is
Δpy = mvfy − mviy = m(vfy − vi
The average y force on the ball is
Fy = Δpy/Δt = Iy/Δt = (52 kg·m/s)/(0.20 s) = 2.6 × 102 N
Since F has no x component, the average force has magnitude 2.6 ×102 N and points in the
y direction (away from the wall).
Problem #3
A child bounces a superball on the sidewalk. The linear impulse delivered by the sidewalk is 2.00N.s during the 1/800 s of contact. What is the magnitude of the average force exerted on the ball by the sidewalk.
Answer:
The magnitude of the change in momentum of (impulse delivered to) the ball is |Δp| = |I| = 2.00 N·s. (The direction of the impulse is upward, since the initial momentum of the ball was downward and the final momentum is upward.)
Since the time over which the force was acting was
Δt = 1/800 s = 1.25 × 10−3 s
then from the definition of average force we get:
|F| = |I|/Δt = 2.00 N.s/(1.25 × 10−3 s) = 1.60 × 103 N
Problem #4
The diagrams below are graphs of Force in kiloNewtons versus time in milliseconds for the motion of a 5kg block moving to the right at 4.0 m/s. (a) What is the magnitude and direction of the impulse acting on the block in each case? (b) What is the magnitude and direction of the average force acting on the block in each case? (c) What is the magnitude and direction of the final velocity of the block in each case?
Answer:
a. Impulse is given by the area under the Ft curves. Since we have simple shapes, it is easy to find the area. For rectangles area is height × base and for triangles area is half the height × base.
i. I = 3 kN × 3 ms = 9 Ns
ii. I = −1 kN × 6 ms = −6 Ns
If the impulse is positive, the net area was above the curve and it is directed to the right, if negative to the left.
b. We know I = FaveΔt where Δt is how long the collision lasts. We read Δt from the graphs, so Fave = I/Δt.
i. Fave = (9 Ns)/(3 ms) = 3000 N
ii. Fave = (−6 Ns)/(6 ms) = 1000 N
If the average force is positive it is directed to the right, if negative to the left. The impulse and force have the same direction
c. Impulse is also equal to the difference in momentum, I = mvf − mvi . We can rearrange our equation for vf , vf = I/m + vi .
i. vf = (9 Ns)/(5.0 kg) + 4 m/s = 5.8 m/s
ii. vf = (6 Ns)/(5.0 kg) + 4 m/s = 2.8 m/s
Problem #5
A machine gun fires 35.0g bullets at a speed of 750.0 m/s. If the gun can fire 200 bullets/ min, what is the average force the shooter must exert to keep the gun from moving?
 |
| Fig.2 |
Answer:
Whoa! Lots of things happening here. Let’s draw a diagram and try to sort things out. Such a picture is given in Fig.2. The gun interacts with the bullets; it exerts a brief, strong force on each of the bullets which in turn exerts an “equal and opposite” force on the gun. The gun’s force changes the bullet’s momentum from zero (as they are initially at rest) to the final value of
pf = mv = (0.0350 kg)(750 m/s) = 26.2 kg·m/s
so this is also the change in momentum for each bullet.
Now, since 200 bullets are fired every minute (60 s), we should count the interaction time as the time to fire one bullet,
Δt = 60 s/200 = 0.30 s
because every 0.30 s, a firing occurs again, and the average force that we compute will be valid for a length of time for which many bullets are fired. So the average force of the gun on the bullets is
Fx = Δpx/Δt = 26.2 kg·m/s/0.30 s = 87.5N
From Newton’s Third Law, there must an average backwards force of the bullets on the gun of magnitude 87.5 N. If there were no other forces acting on the gun, it would accelerate backward! To keep the gun in place, the shooter (or the gun’s mechanical support) must exert a force of 87.5N in the forward direction.
We can also work with the numbers as follows: In one minute, 200 bullets were fired, and a total momentum of
P = (200)(26.2 kg·m/s) = 5.24 × 103 kg·m/s
was imparted to them. So during this time period (60 seconds!) the average force on the whole set of bullets was
Fx = ΔP/Δt = 5.24 × 103 kg·m/s/60.0 s = 87.5N
As before, this is also the average backwards force of the bullets on the gun and the force required to keep the gun in place.
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