A ball of mass m moving with a constant velocity strikes against a ball of same mass at rest. If e = coefficient of restitution, then what will be the ratio of velocity of two balls after collision?

^1-e⁄_1+e
^e-1⁄_e+1
^1+e⁄_1-e
^2+e⁄_e-1

Solution

A ball is dropped from a height h. If the coefficient of restitution be e, then to what height will it rise after jumping twice from the ground?

e h/2
2 e h
e h
e⁴ h

Solution

A bullet of mass ‘a’ and velocity ‘b’is fired into a large block of wood of mass ‘c’. The bullet gets embedded into the block of wood. The final velocity of the system is

\(\frac{b}{a+b}\times c\)
\(\frac{a+b}{c}\times a\)
\(\frac{a}{a+c}\times b\)
\(\frac{a+c}{a}\times b\)

Solution

A bomb of mass 9 kg explodes into the pieces of masses 3 kg and 6 kg. The velocity of mass 3 kg is 16 m/s. The kinetic energy of mass 6 kg in joule is

Solution

Two solid rubber balls A and B having masses 200 & 400 gm respectively are moving in opposite direction with velocity of A equal to 0.3 m/sec. After collision the two balls come to rest when the velocity of B is

0.15 m/sec
1.5 m/sec
–0.15 m/sec
None of these

Solution

A spring of spring constant 5 × 10³ N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

18.75 J
25.00 J
6.25 J
12.50 J

Solution

Consider the following two statement:

I.Linear momentum of a system of particles is zero.

II.Kinetic energy of a system of particles is zero.

Then

I implies II but II does not imply I.
I does not imply II but II implies I.
I implies II and II implies I.
I does not imply II and II does not imply I.

Solution

If \(\overrightarrow{L}\) = 0 ⇒ K.E may or may not be zero.

If K.E = 0 ⇒ \(\overrightarrow{L}\) = 0.

A cord is used to lower vertically a block of mass M, through a distance d at a constant downward acceleration of g/8.Then the work done by the cord on the block is

Mg d/8
3Mg d/8
Mg d
– 7 mg d/8

Solution

TM(g - g/8) = ⁷⁄₈ Mg

Work done by the cord = T × d cos 180º

= ⁷⁄₈ Mg d (-1) =- 7 Mg d / 8.

If the linear momentum is increased by 5%, the kinetic energy will increase by

50%
100%
125%
10%

Solution

A motor cycle is moving along a straight horizontal road with a speed v₀. If the coefficient of friction between the tyres and the road is , the shortest distance in which the car can be stopped is

\(\frac{v_{0}^{2}}{2\mu g}\)
\(\frac{v^{2}}{\mu }\)
\(\left (\frac{v^{0}}{\mu g} \right )^{2}\)
\(\frac{v_{0}}{\mu g}\)

Solution

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