# Calculating new velocity from velocity and momentum? Confusing

Hiya! Sam here,...and I'm new. I need help with this online homework problem. To be honeset, I have no idea where to even begin. I've tried the momentum equation where mass times velocity times a proportionality factor gamma, and I think it's the wrong start. Please help??

Question: A skater with a mass of 85kg is moving with a velocity of <3,1,0> m/s at t=3.2 s. If their momentum changes at a rate of <0, 170, 0> kg m/s^2 until t=3.7s, what is their new velocity?

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Maybe you can use this equation...

$\dot{p}=m\frac{\Delta v}{t}=ma$

CAF123
Gold Member
Hiya! Sam here,...and I'm new. I need help with this online homework problem. To be honeset, I have no idea where to even begin. I've tried the momentum equation where mass times velocity times a proportionality factor gamma, and I think it's the wrong start. Please help??

Question: A skater with a mass of 85kg is moving with a velocity of <3,1,0> m/s at t=3.2 s. If their momentum changes at a rate of <0, 170, 0> kg m/s^2 until t=3.7s, what is their new velocity?
p=γmv is appropriate for relativistic velocities, this is clearly not the case here so you can use the classical form p=mv. It wouldn't be wrong to use the former eqn but γ would just be 1.

Can you write what 'If their momentum changes at a rate of <0, 170, 0> kg m/s^2' means mathematically?

p=γmv is appropriate for relativistic velocities, this is clearly not the case here so you can use the classical form p=mv. It wouldn't be wrong to use the former eqn but γ would just be 1.

Can you write what 'If their momentum changes at a rate of <0, 170, 0> kg m/s^2' means mathematically?
Thank you! I'll try what you have given me.
Unfortunately, I do no quite understand what you mean in your question... CAF123
Gold Member
Thank you! I'll try what you have given me.
Unfortunately, I do no quite understand what you mean in your question... Do you understand the notation <..,...,...>? It is a vector in 3D space, so here the entries represent x, y and z directions. The rate of change of momentum in x and z directions are zero, while that in y is non zero:$$\frac{dp_x}{dt}= \frac{dp_z}{dt}= 0\,\,\,\text{while}\,\,\,\frac{dp_y}{dt} = 170$$ You can use the last equation to get eqn for momentum in y between the two times given.