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1 out of 5 attempts Assistance Check My Work Show Me Practice This Question Print A soccer ball rolls out of a gym through the center of a doorway of the room next door. The adjacent room is 5.42 m by 5.42 m with the 1.28 m wide doorway located at the center of the wall. The ball hits the center of a side wall at 45.0°. If the coefficient of restitution of the soccer ball is 0.850, does the ball bounce back out of the room? (Note the ball rolls without slipping, so no energy is lost to the floor). which (select) enough for the ball to escape the room

Explanation / Answer

form the diagram we can see

the door is in the middle of the wall of the square room

a. the ball comes symmetrically from the center of the door to the wall at angle = 45 deg angle of incidence

now, from the definition of coefficient of restitution

angle of reflection = r

then

v1cos(r) = n*vcos(i)

n = coefficeint of restitution = 0.85

also, from conservation of moementum

v1*sin(r) = v*sin(i)

hence

tan(r) = tan(i)/n

r = 49.63546342 deg

hence

v1 = 0.9280355 v

hence this ball touches the other wall at height y from the horizontal wall in the figure

hence

cos(r) = y/sqrt(y^2 + L^2/4)

L = 5.42 m

hence

0.41944847617114(y^2 + 5.42^2/4) = y^2

y = 2.3035000005 m .. (1)

again

tan(r') = tan(i')/n

i' = 90 - r

hence

r' = 45.00000000 deg

also, v2 = v1*sin(i')/sin(r') = 0.84999994547 v

hence the ball strikes the third wall at distance y from the right corner

hence

cos(r') = y/sqrt(y^2 + (L - 2.3035000005 )^2)

0.5(y^2 + 9.71257224688350000025 ) = y^2

y = 3.1164999 m

again

tan(r") = tan(i")/n

i" = 90 - r' = 45 deg

r" = 49.6354634 deg

v3 = v2*sin(i")/sin(r") = 0.7888301754 v

and

the ball strikes the left side of the wall at distance y from the top corner

cos(r") = y/sqrt(y^2 + (L - 3.1164999 )^2)

0.4194484765(y^2 + 5.30611271070001) = y^2

y = 1.95797508680008 m

now the door ranges form y = 5.42/2 +- 1.28/2 = 2.07 m to 3.3035000005m

and y we obtained does not lie in this range

hence the ball does not leave the room

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