accelerlation is change of velocity/time
= (vf-vi)/time
Now for the details. Let the direction of vi be positive, so vf is negative.
= (-18-28)/time watch the units of time.
Hi,
I honestly don't know how to approach this question and what equation to use.
A 50.0 g Super Ball traveling at 28.0 m/s bounces off a brick wall and rebounds at 18.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)
Can anyone help me? What equation would I use and would I use -g or a +g?
Thanks in advance.
2 answers
Force = rate of change of momentum
= (momentum out - momentum in)/time
momentum out = -50*10^-3*18 kg m/s
momentum in = +50*10^-3*28
momentum out - momentum in = -50*10^-3*46
Force = -50*10^-3*46/3.5*10^-3 s
acceleration = Force/mass = -46/3.5*10^-3
= -13.1*10^3 = 1.31*10^4 m/s^2
= (momentum out - momentum in)/time
momentum out = -50*10^-3*18 kg m/s
momentum in = +50*10^-3*28
momentum out - momentum in = -50*10^-3*46
Force = -50*10^-3*46/3.5*10^-3 s
acceleration = Force/mass = -46/3.5*10^-3
= -13.1*10^3 = 1.31*10^4 m/s^2
1 answer