Work, Power and EnergyHard
Question
A ball loses 15.0% of its kinetic energy when it bounces back from a concreate wall. With what speed you must throe it vertically down from a height of 12.4 m to have it bounce back to the same heigh (ignore air resistance)?
Options
A.6.55m/s
B.12.0m/s
C.8.6m/s
D.4.55m/s
Solution
Given: h = 12.4, v = ?
∴ v2 = u2 = 2gh
i.e.,v2 = u2 + 2 + 9.8 × 12.4
= u2 + 243.04
Kinetic energy of the ball when it just hits the wall
=
mv2 = m(u2 + 243.04)
The K.E. of ball after the impect
=
m(u2 + 243.04)
m(u2 + 243.04)
Let v2 be the upward velocity just after the collision with the ground.
So,
mv22 
m(u2 + 243.04)
v22=
(u2 + 243.04)
Now, taking upward motion
v = 0,u = v2
∴ v2 - 2gh
0 =(u2 + 243.04) - 2 × 9.8 × 12.4
u2 = 36.46
u2 =
= 42.89
u = 6.55m/ s
∴ v2 = u2 = 2gh
i.e.,v2 = u2 + 2 + 9.8 × 12.4
= u2 + 243.04
Kinetic energy of the ball when it just hits the wall
=
mv2 = m(u2 + 243.04)The K.E. of ball after the impect
=
m(u2 + 243.04) m(u2 + 243.04)Let v2 be the upward velocity just after the collision with the ground.
So,
mv22 m(u2 + 243.04)v22=
(u2 + 243.04)Now, taking upward motion
v = 0,u = v2
∴ v2 - 2gh
0 =
(u2 + 243.04) - 2 × 9.8 × 12.4u2 = 36.46 u2 =
= 42.89u = 6.55m/ s
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