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 the 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 kgm/s. At what height was the basketball dropped?

Explanation / Answer

Q1.

impulse applied=change in momentum

==>impulse applied=mass*(final speed-initial speed)

==>39=0.105*(final speed-0)

==>final speed=371.428 m/s

Q2. magnitude of impulse=mass*change in velocity

=0.35*(2.34-(-20.8))=8.1 N.s

Q3.

let i and j are unit vectors along east and north .

then total momentum in vector notation=0.1*(30 i -30 j)

magnitude of total momentum=0.1*sqrt(30^2+30^2)=4.2426 kg.m/s

Q4.

impulse applied=force*time duration=mass*change of momentum

let speed just before collision is v.

then 18000*0.04=61*(v-0)

==>v=11.8 m/s

let height from which he has fallen is h meters.

initial velocity of this fall=0 m/s

final velocity=11.8 m/s

acceleration due to gravity=9.8 m/s^2

using the formula:

final velocity^2-initial velocity^2=2*acceleration*distance

==>11.8^2-0^2=2*9.8*h

==>h=7.104 meters

height from which the student falls is 7.104 meters.

Q5. just before striking the floor , the speed of the ball=momentum/mass

=6.334 m/s

let height be h.

using the formula:

final velocity^2-initial velocity^2=2*acceleration*distance

==>6.334^2-0^2=2*9.8*h

==>h=2.047 meters

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