Question

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction perpendicular to the line between the batter and the pitcher. What is the magnitude of the impulse delivered to the baseball?

A) 9.80 Ns
B) 3.50 Ns
C) 7.00 Ns
D) 5.60 Ns
E) 1.40 Ns

Answer 1

The magnitude of the impulse delivered to the baseball is 7.0 Ns

Step-by-step explanation:

Given;

mass of the foul ball, m = 0.14 kg

initial velocity, u = 40 m/s

final velocity, v = 30 m/s in perpendicular direction

Impulse is given as change in momentum;

initial momentum in horizontal direction, Pi = mu

Pi = 0.14 x 40 = 5.6 Ns

final momentum in perpendicular direction, Pf = mv

Pf = 0.14 x 30

Pf = 4.2 Ns

The resultant impulse is given by;

J² = 5.6² + 4.2²

J² = 49

J = √49

J = 7.0 Ns

Therefore, the magnitude of the impulse delivered to the baseball is 7.0 Ns