Question

A 140 g baseball is moving horizontally to the right at 39 m/s when it is hit by the bat. The ball flies off to the left at 51 m/s , at an angle of 25∘ above the horizontal.

Part A: What is the magnitude of the impulse that the bat delivers to the ball?

Express your answer with the appropriate units.

Part B: What is the direction of the impulse that the bat delivers to the ball?

Express your answer in degrees above the horizontal.

Answer 1

Mass of ball m= 140 g = 0.140 kg

Initial velocity of ball   v₁ = 39 i m/s

Final velocity of ball   v₂ = 51 ( -cos 25⁰i +sin 25⁰  j ) m/s

[ i , j are unit vectors along positive X and positive Y axes respectively.]

Impulse = Change in momentum

Impulse = m (v₂ - v₁ ) = 0.140 ( -51 cos 25⁰i + 51 sin 25⁰j - 39 i ) = (-11.931 i + 3.018 j )

Part A)

Magnitude of impulse = [(-11.931)² +(3.018)² ]^1/2 = 12.3 kg.m/s

Part B)

Direction of impulse is 180⁰+ tan^-1 ( 3.018 / -11.931) = 180⁰ -14.2⁰ = 165.8⁰ counterclockwise with the positive x axis.

Above the horizontal, direction of impulse is tan^-1 ( 3.018 / 11.931)= 14.2⁰