10m/s - -10m/s = at

if a = 10m/s², t=2

f*t = momentum change

Solved

Posted on 2006-04-18

A ball of mass 1.5kg is thrown upward. It leaves the thrower's hand with a velocity of 10m/s. The following questions refer to the motion after the ball has lef tthe thrower's hand. Assume that the upward direction is positive.

a) How long does it take for the ball to return to the throwers hand?

I suppose I could do this problem using simple 1-dimensional kinematics....but i've been learning momentum, impulse etc. So i'm trying to do this problem using my new knowledge :/

The only thing I could think of that involved time was the fact that Impluse is the integral from [t1->t2] of f(x) dx, where f(x) is a force function. So i'm not really sure how to do this part.

b) What is the final velocity of the ball (just before it reaches the hand).

This is easy -v_1.

v_2 = -v_1

c) What is the change in momentum of the ball.

Again fairly easy.

p_2 - p_1 = mv_2 - mv_1 = -mv_1 - mv_1 = -2mv_1 = -30 kg m/s

d) What is the impluse calculated from change in momentum.

J = Change in P, thus...

J = -2mv_1 = -30 N*s

e) What is the average force acting on the ball

I think this might tie in to part (a), but i'm not sure.

-Brian

a) How long does it take for the ball to return to the throwers hand?

I suppose I could do this problem using simple 1-dimensional kinematics....but i've been learning momentum, impulse etc. So i'm trying to do this problem using my new knowledge :/

The only thing I could think of that involved time was the fact that Impluse is the integral from [t1->t2] of f(x) dx, where f(x) is a force function. So i'm not really sure how to do this part.

b) What is the final velocity of the ball (just before it reaches the hand).

This is easy -v_1.

v_2 = -v_1

c) What is the change in momentum of the ball.

Again fairly easy.

p_2 - p_1 = mv_2 - mv_1 = -mv_1 - mv_1 = -2mv_1 = -30 kg m/s

d) What is the impluse calculated from change in momentum.

J = Change in P, thus...

J = -2mv_1 = -30 N*s

e) What is the average force acting on the ball

I think this might tie in to part (a), but i'm not sure.

-Brian

6 Comments

10m/s - -10m/s = at

if a = 10m/s², t=2

f*t = momentum change

Also, the question goes on to ask the following...

After the ball hits the throwers hand it comes to rest in 0.25s

What is the net implus exerted on the ball?

J = change in p

= -mv_1

= -(1.5kg)(-10m/s) = 15 N*s

What is the average force exerted by the hand on the ball (dont forget gravitational force)

J = F_net * t

J = (F_h + F_e)t

J = F_h*t + mgt

J - mgt

--------- = F_h

t

F_h = (15N*s + (1.5kg)(9.8m/s^2)(.25s))/.

Again doesnt seem resonable, is this work correct ozo?

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