Momentum: Jumping on a trampoline

[smagro]

Homework Statement

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.

Homework Equations

Momentum = mass x velocity

Force = change in momentum/time

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 I am momentum = final momentum - initial momentum
= -228kgm/s - 252kgm/s = -480kgm/s

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

Is this correct?

 

[Doc Al]

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

 

[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

what is your opinion?

 

[smagro]

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

 

[Doc Al]

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

what is your opinion?

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.

 

[Chestermiller]

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

what is your opinion?

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

Chet

 

[Doc Al]

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

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.

 

[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!

 

[Doc Al]

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!

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.)

 

[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.

 

[Doc Al]

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

Most definitely!