Bouncing Ball Example
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- 0.5 s
- 1 s
- 1.5 s
- 2 s
- √2hg(1+e1−e)
- √hg(2−e2+e)
- √2gh(2−e2+e)
- √hg(e1−e)
- 10 m
- 4.9 m
- 9.8 m
- None of these
- (1−e2)u2sin2θg
- e2u2sin2θg
- (1−e)u2sinθcosθg
- (1+e)u2sin2θg
A steel ball falls from a height ‘h’ on a floor for which the coefficient of restitution is e. The height attained by the ball after the first rebounds is
- he2
- h/e
- h/e2
- he
- √2hg
- 2√2hg
- 3√2hg
- 4√2hg
- 1
- 0
- 12
- √32
- 1.5 s
- 2.0 s
- 3.0 s
- 1.0 s
A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizontal frictionless surface.When the spring is 4.0 cm shorter than its equilibrium length, the speed of 17^1/2m/s.The greatest speed of the block is
1)4m/s
2)6m/s
3)5m/s
4)9m/s
5)3m/s
- The total time taken in moving from O to C is ae2u(e2+e+1)
- The free range on the horizontal plane b=2uvg
- be2=a(e2+e+1)
- All of these
What is the percentage change in the momentum of the body, if the mass of a body is doubled and its velocity is reduced by half?
0 %
10 %
50 %
100 %
- 2√2×103m/s
- 4√2×103m/s
- 4×103m/s
- 4√3×103m/s
- R′=eR
- R′=e2R
- R′=R
- R′=Re
A ball is dropped from a height of . If the coefficient of restitution is , what will be the height attained after the first bounce?
A ball dropped freely takes 0.2 seconds to cross the last 6 m distance before hitting the ground. total time of fall is ? ( g =10m/s2 )
2.9s
3.1s
2.7s
0.2s
- 1
- 34
- 12
- 14
- m√2gh(ee(1+e))
- √2mgh(e1−e)
- m√2gh(2e1+e)
- m√2gh(1+e1−e)
- VC = VB = VA
- VC = VA≠VB
- VC = VB≠VA
- VC≠VB≠VA
A body falls from a height h on a horizontal surface and rebounds. Then it falls again and rebounds and so on. If the restitution coefficient is 13, the total distance covered by the body before it comes to rest is
5h4
3h
h4
2h
A ball is dropped from a hieght of 5m onto a Sandy floor and penetrates the sand upto 10 cm before coming to rest . Find the retardation of the ball in sand assuming it to be uniform.
A ball is dropped from a height. If it takes 0.2s to cross the last 6 m before hitting the ground, find the height from which it was dropped.
- 72
- 60
- 36
- 20
- 2 m/s
- 4.2 m/s
- 6 m/s
- 11.66 m/s
- 0.97 cm
- 1.96 cm
- 0.67 cm
- 4.67 cm
- √2hg(1+e1−e)
- √hg(2−e2+e)
- √2gh(2−e2+e)
- √hg(e1−e)
(a) Prove that when the system is in equilibrium, separation between the balls is x=(q2ℓ2πε0mg)1/3, assuming the angle made by the threads with the vertical line passing through the point of suspension is small.
(b) Specify the rate (dq/dt) with which the charge from each sphere leaks off if their velocity of approach varies as v=a√x, where a is a constant.