A model rocket is constructed with a motor that can provide a total impulse of 3

Question

A model rocket is constructed with a motor that can provide a total impulse of 39.0 Ns. The mass of the rocket is 0.105 kg. What is the speed that this rocket achieves when it Is launched from rest? Neglect the effects of gravity and air resistance. A volleyball is spiked so that its incoming velocity of +2.34 m/s is changed to an outgoing velocity of -20.8 m/s. The mass of the volleyball is 0.350 kg. What is the magnitude of the impulse that the player applies to the ball? Two arrows are fired simultaneously with the same speed of 30.0 m/s. Each arrow has a mass of 0.100 kg. One is fired due east and the other due south. What is the magnitude of the total momentum of this two-arrow system? A student (m = 61 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.04 s. The average force exerted on him by tin- ground is +18000 N. where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground. A basketball (m 0.60 kg) is dropped from rest. Just before striking the floor, the ball has a momentum whose magnitude is 3.8 kg m/s. At what height was the basketball dropped?

Explanation / Answer

2.The impulse is the change of momentum, which is the mass times the change in velocity

(outgoing vel - incoming vel)
= 0.350(-20.8 - 2.34) = -8.099 kg m/s and the magnitude is 8.099 kg m/s.

3.p1 = m*v = .1*30 = 3 kgm/s
P = p1e + p1s = 3*2 = 4.24 kgm/s @ 45°

4. first you need to find the equivalent acceleration that would cause that force:

a = F/m = 295.08 m/s^2
Next, use the equation:
V = at with t=.04 to find the velocity
V = 295.08 (.04) = 11.80 m/s
now find the height that would cause that velocity using:
V^2 = 2ax
11.80^2 = 2(9.8)x
so x = 7.10 m

5.

p = mv, or momentum = mass x velocity. So v = p/m, or velocity is momentum divided by mass.
v = (3.8 kg m/s)/(0.6 kg) = 6.33 m/s. So the ball hits the floor at a speed of 6.33 m/s. How far did it have to fall to hit the floor at that distance? We can work this out using one of the four kinematic equations of motion. Since we are considering downwards motion only, let's define downward to be the positive direction. What information have we got?

u = initial velocity = 0
v = final velocity = 6.33 m/s
a = acceleration due to gravity = 9.8 m/s²
d = distance = ?
v² = u² + 2ad. And since u = 0,
v² = 2ad. Rearrange to find d
d = v²/2a = 6.33²/2*9.8 = 2.04 m

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