Question

A 7 kg steel ball strikes a wall with a speed
of 8.77 m/s at an angle of 59.4  with the
normal to the wall. It bounces off with the
same speed and angle

If the ball is in contact with the wall for
0.237 s, what is the magnitude of the average
force exerted on the ball by the wall?
Answer in units of N

Answer 1

the force exerted by the wall will be reaction force in the normal direction from the wall exerted by the ball

Answer 2

the idea is to make the velocity (8.77 m/s) and mass (kg) into a force, using the time given.
remember that
a force = a newton = m* a = kg*(m/s^2)
we have kg, and m and s and s again. so we will set it up like this:
\[mass * velocity * \frac{ 1 }{ time } = kg*\frac{ meters }{ \sec } * \frac{ 1 }{ \sec } = \frac{ kg*m }{ s^2 } = m*a = F\]

so plug in the numbers you are given in the the middle equation i gave so it now looks like this:

\[7kg * \frac{ 8.77m }{ s } * \frac{ 1 }{ 0.237s } = F = 259\]

we now have a force that the ball is putting on the wall, from here you will sum your forces. break the force of the ball into x and y components using the angle it impacts the wall at. then you can solve for the magnitude of the normal force of the wall.