A 2.1 kg steel ball strikes a massive wall at 13.2 m/s at an angle of 64.8 ◦ with the perpendicular to the plane of the wall. It bounces off the wall with the same speed and angle it originally had:__________.

going through the question you would notice that some detail is missing, using search engines u was able to find a similar question on "https://socratic.org/questions/a-2-3-kg-steel-ball-strikes-a-wall-with-a-speed-of-8-5-m-s-at-an-angle-of-64-wit"

and here is the question i found

"A 2.3 kg steel ball strikes a wall with a speed of 8.5 m/s at an angle of 64⁰ with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.448 s. What is the average force exerted by the ball?"

you would notice that there is a change in the values from the question posted, hence we would only take the following part to complete our question, " If the ball is in contact with the wall for 0.448 s what is the average force exerted by the ball?" while retaining all original detail.

solution

mass of ball (m) = 2.1 kg

speed of ball (v) = 13.2 m/s

angle of contact (p) = 64.8°

time of contact (t) = 0.448 s

What is the average force exerted by the ball?

The average force exerted by the ball = [tex]\frac{change in momentum}{change in time}[/tex]

where

The momentum changes only the direction perpendicular to the wall, hence the component of momentum perpendicular to the wall = m x v x sin (p) = 2.1 x 13.2 x sin 64.8 = 25.1 kg.m/s

since the ball strikes the wall and bounces off it with the same speed

and at the same angle, the component of momentum acting

perpendicular to the wall remains the same while hitting and leaving

the wall but in opposite directions.

Hence the component of momentum acting perpendicular to the wall while hitting and leaving the wall will be 25.1 kg.m/s and -25.1 kg.m/s respectively.

change in momentum = 25.1 - (-25.1) = 25.1 + 25.1 = 50.2 kg.m/s

change in time = 0.448 s

now substituting the above into the equation we have

The average force exerted by the ball = [tex]\frac{50.2}{0.448}[/tex] = 112 N