During a neighborhood baseball game in a vacant lot, a particularly wild hit sen

Question

During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.145-kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 15.3 m / s, shatters the glass as it passes through, and leaves the window at 11.9 m / s with no change of direction.

What is the direction of the impulse that the glass imparts to the baseball? Calculate the magnitude of this impulse (a positive number).

The ball is in contact with the glass for 0.0105 s as it passes through. Find the magnitude of the average force of the glass on the ball (a positive number).

Explanation / Answer

Here ,

m 0.145 Kg

v1 = 15.3 m/s

v2 = 11.9 m/s

as the velocity of the ball is decreasing , the direction of impulse is oposite to the direction of motion of the baseball.

magnitude of impulse = change in momentum

magnitude of impulse = 0.145 * (15.3 - 11.9)

magnitude of impulse = 0.493 Kg.m/s

the magnitude of impulse is 0.493 Kg.m/s

----

magnitude of average force = impulse/time

magnitude of average force = 0.493/.0105 N

magnitude of average force = 47 N

the magnitude of average force on the ball is 47 N

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.