Re: Momentum, Mechanics 1

(a) you've calculated the correct value, but note that Newtons are units of force, not momentum.

(b) conservation of momentum ...

Jason ...

$\displaystyle (125 \, kg)(3 \, m/s) = (120 \, kg)v_f + (5 \, kg)(9 \, m/s)$

Keith ...

$\displaystyle (120 \, kg)(3 \, m/s) + (5 \, kg)(9 \, m/s) = (125 \, kg)v_f$

Re: Momentum, Mechanics 1

Initially, Jason has momentum of 125*3= 375 "kg-m/s", not "Newtons", as skeeter said. If he throws the ball forward, with velocity v (relative to the ground), The ball has momentum 5*9= 45 kg-m/s and, by conservation of momentum, Jason has momentum 375- 45= 330 and so his new velocity is 330/120 = 2.75 m/s.

Keith's initial momentum was 120*3= 360 kg-m/s. After he catches the ball his momentum will be 360+ 45= 405 kg-m/s and the total mass of Keith, truck, and ball is 125 kg so his new velocity is 405/125= 3.24 m/s.

Of course, the problem asks for the difference in their speeds.

Re: Momentum, Mechanics 1

Thank you Skeeter and Halls of Ivy for your time and detailed explanation. Thank you Thank you Thank you