# Momentum: Jumping on a trampoline

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1. Apr 16, 2015

### smagro

1. The problem statement, all variables and given/known data

A gymnast of mass 40kg is practising on a trampoline.
The gymnast lands with a speed of 6.3m/s. The gymnast rebounds with a speed of 5.7m/s.
a) Calculate the change in momentum of the gymnast
b) The gymnast was in contact with the trampoline for 0.50s. Calculate the average force exerted by the trampoline on the gymnast.

2. Relevant equations
Momentum = mass x velocity

Force = change in momentum/time

3. The attempt at a solution

a) momentum before = 40kg x 6.3m/s = 252kgm/s

momentum after = 40kg x (-5.7m/s) = -228kgm/s

change im momentum = final momentum - initial momentum
= -228kgm/s - 252kgm/s = -480kgm/s

b) Force = -480 / 0.50s = -960N.

Is this correct?

2. Apr 16, 2015

### Staff: Mentor

Looks good to me! (You have used the convention that down = positive, which is perfectly OK.)

3. Apr 16, 2015

### smagro

some arguments against my working were that the gymnast change in momentum was from 0m/s to 5.7m/s because the gymnast was at rest at the maximum stretching point of the trampoline.

this will make the change in momentum be 228kgm/s - 0 kgm/s = 228kgm/s.

then the force is 228/0.50 = 456N

4. Apr 16, 2015

### smagro

is it also good to argue that the upward force by the trampoline should be greater than the weight of gymnast?

5. Apr 16, 2015

### Staff: Mentor

I would say that that is wrong. Note that if you only count from the lowest point (where the gymnast is momentarily at rest) the time of contact time would be less than 0.50 seconds.

6. Apr 16, 2015

### Staff: Mentor

This is not a matter of opinion. These arguments are incorrect.

Chet

7. Apr 16, 2015

### Staff: Mentor

It had better be! Otherwise, how can the gymnast be accelerated upwards?

Also, what you've calculated is the net upward force on the gymnast. You need to factor out gravity to get the force of the trampoline. (Sorry for not pointing that out!)

So you'll need to correct your initial analysis.

8. Apr 16, 2015

### smagro

what if the collision was an elastic collision. momentum before should be equal to the momentum after.

so the gymnast should rebound with a speed of 6m/s.

but if we work momentum, momentum (being a vector quantity), we get 40kg x 6.3m/s = 252kg/ms

momentum after = 40kg x -6.3m/s = -252kgm/s.

So momentum is not conserved!!!

9. Apr 16, 2015

### Staff: Mentor

Why would you think momentum is conserved? There's an external force acting on the gymnast--the trampoline!

(Don't confuse this with the usual problem of two objects colliding, such as two balls. There, the total momentum of both objects is conserved.)

10. Apr 16, 2015

### smagro

thanks....

summary: momentum is a vector quantity and direction has to be considered in working out the change in momentum.

There is no argument about that.

11. Apr 16, 2015

### Staff: Mentor

Most definitely!