A batter hits a foul ball. The 0.14-kg baseball that was approaching him at 40 m/s leaves the bat at 30 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?

Question

A batter hits a foul ball. The  0.14-kg baseball that was approaching him at  40 m/s leaves the bat at  30 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?

Quizplus · Accepted Answer

Impulse is the change in momentum. The initial momentum of the baseball is:$$p_i = m v_i = (0.14 \mathrm{~kg}) (40 \mathrm{~m/s}) = 5.6 \mathrm{~kg \cdot m/s}$$The final momentum of the baseball is:$$p_f = m v_f = (0.14 \mathrm{~kg}) (30 \mathrm{~m/s}) = 4.2 \mathrm{~kg \cdot m/s}$$The impulse delivered to the baseball is:$$J = p_f - p_i = 4.2 \mathrm{~kg \cdot m/s} - 5.6 \mathrm{~kg \cdot m/s} = -1.4 \mathrm{~kg \cdot m/s}$$The magnitude of the impulse is:$$|J| = |-1.4 \mathrm{~kg \cdot m/s}| = 1.4 \mathrm{~N \cdot s}$$Therefore, the correct answer is A.

A Batter Hits a Foul Ball 7.0 N⋅s7.0 \mathrm {~N} \cdot \mathrm { s }7.0 N⋅s

A Batter Hits a Foul Ball $7.0 \mathrm {~N} \cdot \mathrm { s }$