Question

A ball with an initial velocity of 10 m/s moves at an angle 60 0 above the + x -direction. The ball hits a vertical wall and bounces off so that it is moving 60 0 above the − x -direction with the same speed. What is the impulse delivered by the wall?

Answer 1

The mass of the ball is missing, So, a mass of 9 kg was assigned to it, based on a similar question obtained online with a few parameters changed.

Answer:

90 N.s

Explanation:

The image showing the scenario described in the question is attached to this solution.

According to the Newton's second law of motion,

Impulse = Magnitude of the Change in momentum.

Momentum before collision = mv

m = 9 kg

v = (10 cos 60î + 10 sin 60j) m/s

v = (5î + 8.66j) m/s

Momentum before collision = 9(5î + 8.66j) = (45î + 77.94j) kgm/s

Momentum after collision = mv

v = (-10 cos 60î + 10 sin 60j) m/s

v = (-5î + 8.66j) m/s

Momentum after collision = 9(-5î + 8.66j) = (-45î + 77.94j) kgm/s

Change in momentum = (Momentum after collision) - (Momentum before collision)

Change in momentum = (-45î + 77.94j) - (45î + 77.94j) = (-90î) kgm/s

Magnitude of the change in momentum = √[(-90)²] = 90 kgm/s

Magnitude of the Impulse delivered by the wall = Magnitude of the change in momentum of the ball = 90 kgm/s = 90 N.s

Hope this Helps!!!