Question

During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.146 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 10.7 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 0.0106 s as it passes through. Find the magnitude of the average force of the glass on the ball (a positive number).

Answer 1

Answer:

Impulse, |J| = 0.6716 kg-m/s

Force, F = 63.35 N

Explanation:

It is given that,

Mass of the baseball, m = 0.146 kg

Initial speed of the ball, u = 15.3 m/s

Final speed of the ball, v = 10.7 m/s

To find,

(a) The magnitude of this impulse.

(b) The magnitude of the average force of the glass on the ball.

Solution,

(a) Impulse of an object is equal to the change in its momentum. It is given by :

$J=m(v-u)$

$J=0.146\ kg(10.7-15.3)\ m/s$

J = -0.6716 kg-m/s

or

|J| = 0.6716 kg-m/s

(b) Another definition of impulse is given by the product of force and time of contact.

t = 0.0106 s

$J=F\times \Delta t$

$F=\dfrac{J}{\Delta t}$

$F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}$

F = 63.35 N

Hence, this is the required solution.