Hi,
A squash ball of mass 0.035 kg hits the wall at speed of 78 m/s and
rebounds straight back at 61 m/s. The ball is in contact with the wall for
0.44 s.
a) What is the original momentum of the squash ball? 2.73 kg-m/s
b) Calculate the impulse imparted to the ball by the wall.
deltap=mdeltav
deltap=0.035kg*17m/s= 0.595kg-m/s
c) What force acted on the squash ball from its impact?
deltap/t=f
0.595kg-m/s/0.44s=1.35N
I'm fairly certain b & c are wrong but I don't know what to change...
Thanks for your help
b. impulse=changemomentum= m(vf-vi) now remember, then the change of velocity is in opposite directions, so impulse=m(61+78)
c. force= impulseAbove/timeonWall
b) impulse is the change in momentum ... final minus initial
... look at your deltav value ... velocity is a vector ... magnitude AND direction
this should also "fix" c)
So to make sure I understand...
Because it's bouncing back off the wall & change in velocity is now negative I should have used impulse=m(-change in velocity)
impulse=0.035kg*-(-61-78)
impulse=0.035kg*139m/s
impulse=4.9 kg-m/s
Is that right?
Thanks!
yes.
Let's go through the calculations again to find the correct answers for parts (b) and (c).
b) To calculate the impulse imparted to the ball by the wall, we need to use the formula:
Impulse = Change in momentum
The change in momentum can be calculated by subtracting the final momentum from the initial momentum:
Change in momentum = Final momentum - Initial momentum
Given that the initial momentum is 2.73 kg·m/s and the final momentum is unknown, we can rewrite the equation as:
Impulse = Final momentum - 2.73 kg·m/s
Note that the impulse is also equal to the force multiplied by the time of contact:
Impulse = Force × Time
Since we have the time of contact as 0.44 s, we can rearrange the equation to solve for the force:
Force = Impulse / Time
Now, let's calculate the impulse:
Impulse = 0.035 kg × (61 m/s - 78 m/s)
Impulse = 0.035 kg × (-17 m/s)
Impulse = -0.595 kg·m/s
Now that we have the impulse, let's calculate the force:
Force = -0.595 kg·m/s / 0.44 s
Force ≈ -1.35 N
It's important to note that the negative sign indicates that the force exerted on the ball by the wall is in the opposite direction (opposite to the direction of motion).
c) The force exerted on the squash ball from its impact is approximately 1.35 N in magnitude and it acts on the ball in the opposite direction to its motion.
So the corrected answers are:
b) The impulse imparted to the ball by the wall is approximately -0.595 kg·m/s.
c) The force acted on the squash ball from its impact is approximately -1.35 N.