While in horizontal flight at a speed of 20 m/s,a baseball of mass 0.11 kg is struck by a bat.After leaving the bat,the baseball has a speed of 29 m/s in a direction opposite to its original direction.The magnitude of the impulse given the ball is
A)0.99 kg · m/s
B)5.4 kg · m/s
C)2.2 kg · m/s
D)3.2 kg · m/s
E)0.55 kg · m/s

Question

Quizplus · Accepted Answer

We can use the impulse-momentum theorem which states that the impulse given to an object is equal to its change in momentum. Mathematically:

I = Δp

Where I is the impulse and Δp is the change in momentum.

The initial momentum of the ball is:

p1 = m*v1

Where m is the mass of the ball, and v1 is its initial velocity.

p1 = (0.11 kg)(20 m/s) = 2.2 kg·m/s

The final momentum of the ball is:

p2 = m*v2

Where v2 is the final velocity of the ball after being struck by the bat. Note that the direction of the velocity is opposite to the initial direction.

p2 = (0.11 kg)(−29 m/s) = −3.19 kg·m/s

The change in momentum is:

Δp = p2 − p1 = −5.39 kg·m/s

The magnitude of the impulse is the absolute value of the change in momentum:

|I| = |Δp| = 5.39 kg·m/s

Therefore, the answer is B) 5.4 kg·m/s.

While in Horizontal Flight at a Speed of 20 M/s,a