Solving an Acceleration Problem Involving a Super Ball

[cmkc109]

Homework Statement

A 55.0 g Super Ball traveling at 30.0 m/s bounces off a brick wall and rebounds at 21.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 5.00 ms, what is the magnitude of the average acceleration of the ball during this time interval?

The Attempt at a Solution

I know how to solve, but don't understand why is it solve this way.

F = m a

F = m dv/dt

Fdt = mdv

F(5E-3) = (55E-3)(21-(-30))

F = 561 N

a = F/m = 561/55E-3

a = 10200 m/s²


I highlighted the part that I don't understand
v1 = 25m/s , but when it rebounces , wouldn't the velocity changes then vf should be different in this case from a ball that is just going straight without hitting a barrier.
Can someone explain it to me? thanks!

 

[haruspex]

It changed from 30 m/s in one direction to 21 m/s in the opposite direction. What is the magnitude of the change in velocity?

 

[cmkc109]

+30m/s and -21m/s

30-(-21) = 51

 

[haruspex]

Right, so what's left that you do not understand. You didn't highlight Fdt = mdv.

 

[cmkc109]

let's say if the ball starts at v1 = 10m/s m and it bounces off the wall, and to another wall, and v2 = 20m/s the avg velocity is still 20-10 ?
and to find acceleration, u do a= v/t , t will be the change in time.
sometimes i don't know when do you need to use the kinematic equations and when do you use use a= v/t or d = v t .
can u explain it please? Thank you very much.

 

[haruspex]

let's say if the ball starts at v1 = 10m/s m and it bounces off the wall, and to another wall, and v2 = 20m/s the avg velocity is still 20-10 ?

The two walls are facing each other? How did its speed get up to 20m/s? Is that after bouncing off two walls? Why do you ask about average average velocity? It was change in velocity that was wanted.

and to find acceleration, u do a= v/t , t will be the change in time.
sometimes i don't know when do you need to use the kinematic equations and when do you use use a= v/t or d = v t .

Typically you are armed with a set of equations valid in a certain context (like, uniform acceleration). You have a context in which they should be valid, and certain knowns, together with one or more unknowns that you wish to calculate.So you think through your equations and see which one has your desired unknown, the rest of its variables being knows. If you have a change in speed and a time over which it occurs you can calculate the average acceleration.

 

[cmkc109]

oh right, the velocity should decrease after it bounces off the wall. i was just making random numbers up. it's just a = v/t , and avg velocity can be found from v2-v1/ t2-t1, and t2-t1 is the change in time..then technically it is looking for average velocity..?
but yea, i mixed up change in velocity as avg velocity..

so for a= v/t , the velocty is not changing but constant? if it is changing , there will be v1 and v2 and will use the kinematics equations?

 

[haruspex]

it's just a = v/t , and avg velocity can be found from v2-v1/ t2-t1, and t2-t1 is the change in time

No, (v2-v1)/ (t2-t1) gives average acceleration, not average velocity. Average velocity would use change in distance: (s2-s1)/(t2-t1)

so for a= v/t , the velocty is not changing but constant? if it is changing , there will be v1 and v2 and will use the kinematics equations?

If the acceleration is constant, you can use (v2-v1)/ (t2-t1) to find it. If it is not constant that formula will give the average acceleration. In the special case where v1=0 and you start timing at t1=0 that reduces to a = v/t.

 

[cmkc109]

thanks!