Seconds uhile it is councaumingto a shop42.) A 0.050 kq white ball is thrown hori zontalytou ordThe bartter suings the bat and hits the ballyboatter of12.3 mls oaapplyingHhe ball to head back in thean ave rage force f 15.714N to te balle oppostecausingdiredtion Causu from the ba Her) at d.70 mis, lossumpgthat away drom hatter is the possitive t-diedin,determine the time that the boot and ball wirein Cantact, Reportin Mondreths ofansaeryoursecond enter humerical volve

42.) A 0.050 kg whildle ball is thrown hori zantally batter of tow ord The boter 12.3 mls a a the bat and hits the ballo suings an ave cage force f 15.714 N to the ball applying the ball to head back in the fausing diredion Cauou from the beather) at a.70 mis, Lossumpg e opposte away Rrom botter is the posstive determine the time that in Centact, Report that the bot and balll uere i hendretns of o ansaer your second ente humerical volve

42.) A 0.050 kg whildle ball is thrown hori zantally batter of tow ord The boter 12.3 mls a a the bat and hits the ballo suings an ave cage force f 15.714 N to the ball applying the ball to head back in the fausing diredion Cauou from the beather) at a.70 mis, Lossumpg e opposte away Rrom botter is the posstive determine the time that in Centact, Report that the bot and balll uere i hendretns of o ansaer your second ente humerical volve

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Question

Seconds uhile it is coun
cauming
to a shop
42.) A 0.050 kq white ball is thrown hori zontaly
tou ord
The bartter suings the bat and hits the bally
boatter of
12.3 mls o
a
applying
Hhe ball to head back in the
an ave rage force f 15.714N to te ball
e opposte
causing
diredtion Causu from the ba Her) at d.70 mis, lossumpg
that away drom hatter is the possitive t-diedin,
determine the time that the boot and ball wire
in Cantact, Report
in Mondreths of
ansaer
your
second enter humerical volve

Expert Solution

Step 1

In this case, an impulse is acted on the ball, which is the case of the change of momentum of the ball in a short time interval due to a large force acting on it.

The impulse on the ball can be found by the change in the momentum of the ball, as

$I = p_{2} - p_{1} p_{2} i s t h e f i n a l a n d p_{1} i s t h e i n i t i a l m o m e n t u m o f t h e b a l l A s t h e b a l l t r a v e l s i n t h e o p p o s i t e d i r e c t i o n a f t e r b e i n g h i t, s o t h e m o m e n t a o f t h e b a l l a r e o p p o s i t e l y d i r e c t e d I = p_{2} - (- p_{1}) I = p_{2} + p_{1} I = m v_{2} + m v_{1} I = m (v_{2} + v_{1}) I = 0.05 (9.7 + 12.3) I = 1.1 N s$

This is the impulse acting on the ball

Step 2

The impulse is also defined as the amount of force acting on the ball in a short time interval t.

So, impulse can also be written as

$I = F ∆ t ∆ t i s t h e t i m e i n t e r v a l f o r w h i c h t h e f o r c e h a s b e e n a c t e d u p o n . S o, ∆ t = \frac{I}{F} ∆ t = \frac{1.1}{15.714} ∆ t = 0.07 s ∆ t = 7 \times 10^{- 2} s$

This is the time for which the bat and ball were in contact.

Step by step

Solved in 3 steps

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