A ball of mass 1.5 kg is thrown upward. It leaves the thrower\'s hand with a vel

Question

A ball of mass 1.5 kg is thrown upward. It leaves the thrower's hand with a velocity of 10 m/s. The following questions refer to the motion after the ball leaves the thrower's hand. Assume that the upward direction is positive. Please show me the work, I am lost, thank you
A. How long before the ball returns to the throwers hand?
B. What is final velocity of the ball just before it reaches the hand?
C. What is the change in momentum of the ball?
D. What is the impulse calculated from the change of momentum?
E. from the impulse, find the average force acting on the ball. What causes the force?

After the ball hits the thrower's hand it comes to a rest in .25 sec.
A. What is the net impulse exerted on the ball by the hand?
B. What is the average force exerted by the hand on the ball during the .25sec that the ball is being stopped?
C.What is the average force exerted by the ball on the hand during this time?

Explanation / Answer

a. We use the equation of motion to find the time it takes for the ball to reach its maximum height. Due to the symmetry of the problem, we can double this to find the amount of time it took to reach back to his hand. v0 = 10m/s, v1 = 0 m/s, g = -9.8 m/s2, t is what we're trying to find

v1 = v0 + g t = gt

t = v1 / g = (10m/s)/(9.8 m/s2) = 1.02 seconds

We now double this to find how long it took to return to his hand:

1.02 seconds * 2 = 2.04 seconds

b. Due to the symetry of the problem, the ball will be traveling -v0 when it returns to his hand. So the ball is traveling -10 m/s. If you would like, you can calculate this using v1 = v0 + at.

c.This problem asks for the change in momentum during the balls flight in the air. v1 in this case refers to the initial 10 m/s, while v0 refers to the -10m/s we deduced in part b.

?p = p1 - p0 = m(v1 - v0) = 1.5kg(-10 m/s - 10 m/s) = -30 kg m/s

The answer is negative because the direction of the ball is opposite at the end of its arc. The answer might be double what you would expect, but the ball had a change in momentum from v1 -> -v1 which is a change in velocity of 2v1

d. We assume a constant force acted on the ball. Impulse is just the change in momentum.

I = ?p = p1 - p0 = m(v1 - v0) = 1.5kg(-10 m/s - 10 m/s) = -30 kg m/s

e. Another way of writing impulse is:

I = F ?t

We can solve for F, plugging in the time from part a and the impulse from part d

F = I/?t = (-30 kg m/s)/(2.04 seconds) = -14.71 kg m/s2 = -14.7 N

We should not that the weight of the object (the force of gravity on the object) is given by:

Fgravity = m g = 1.5 kg (-9.8 m/s2) = -14.7 N

And gravity is the only force acting on the ball, so the force is gravity.

- - - - - - - - - - - -

a. Net impulse is the change in momentum, so we use m = mass = 1.5 kg, v0 is the speed just before it touches his hand, and v1 is the speed once he stops the ball (0m/s):

I = ?p = m(v1 - v0) = 1.5kg(0 m/s - 10 m/s) = 15 kg m/s

b. Again we use another equation of impulse, using the impulse we found in part a and the time given in the problem:

I = F?t

F = I/?t = (15kg m/s)/(.25 seconds) = 60 N

c. Using Newton's third law, the ball will have a force on the hand equal and opposite in sign:

Fhand = -Fball = -60N

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