A 4.20 kg steel ball strikes a wall with a speed of 12.0 m/s at an...

A 4.20 kg steel ball strikes a wall with a speed of 12.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.200 s, what is the average force exerted by the wall on the ball? (Assume right is the positive direction.)

Expert Solution

The figure shows a steel ball coming and striking the surface, making an angle 60^o with the surface. p_x and p_y shows the initial momentum along X and Y direction whereas p_x' and p_y' shows the final momentum along X and Y direction of the steel ball.

It is clear from the figure that there no change in momentum along the Y axis before and after collision. However, along the X axis momentum direction changes but its magnitude is the same according to question. Therefore, Force (F) acts along X axis only given by:

Here, m is the mass of the steel ball

v is velocity of ball whose magnitude is same before and after collision as given in the question "bounces off with the same speed and angle".

We have considered the right side as positive as stated in the question.

The negative sign in the question signifies that the force acts along the negative X axis on the ball.

genius_generous answered 2 years ago

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