Q. 53

Expert-verifiedFound in: Page 108

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A 35 g steel ball is held by a ceiling-mounted electromagnet 3.5 m above the floor. A compressed-air cannon sits on the floor, 4.0 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.0 m above the floor. What was the launch speed of the plastic ball?

The launch speed of the ball is .

The vertical height of the electromagnet is , the horizontal distance of the electromagnet from the canon is . The collision occurs at vertical height of above the floor. The masses of the plastic ball and the steel ball are and respectively.

The formula to calculate the horizontal distance travelled by the plastic ball is given by

Here, are the launch speed, the projection angle and the time taken by the plastic ball respectively.

Substitute for into equation (1) to obtain the relation.

The formula to calculate the final vertical position of the plastic ball is given by

Substitute for , for and for into equation (5) and solve to calculate the time taken by the plastic ball to collide with the steel ball.

Substitute for into equation (2) and simplify to obtain the square of the cosine of the projection angle.

Substitute for and for into equation (4) and simplify to obtain the square of the sine of the projection angle.

Add equation (6) and equation (7) and solve to calculate the required launch speed.

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