A bag of mass M hangs by a long thread and a bullet (massm) comes horizontally with velocity V and gets caught in the bag. Then for the combined (bag + bullet) system

momentum =\(\frac{mvM}{M+m}\)
kinetic energy =\(\frac{mV^{2}}{2}\)
momentum =\(\frac{mV(M+m)}{M}\)
kinetic energy =\(\frac{m^{2}v^{2}}{2(M+m)}\)

Solution

A body is attached to the lower end of a vertical helical spring and it is gradually lowered to its equilibrium position.This stretches the spring by a length x. If the same body attached to the same spring is allowed to fall suddenly,what would be the maximum stretching in this case?

Solution

When body is lowered gradually, its weight acts at C.G. of the spring. When same body is allowed to fall freely,the same weight acts at lower end of the spring.In the latter case, original length (L) of spring is double.As ΔL ∝ L, therefore, ΔL becomes twice in second case i.e. 2x.

Given that a force \(\hat{F}\) acts on a body for time t, and displaces the body by \(\hat{F}\). In which of the following cases, the speed of the body must not increase?

F > d
F < d
\(\hat{F}=\hat{d}\)
\(\hat{F}\perp \hat{d}\)

Solution

Velocity will increase when force is along the direction of displacement i.e.\(\hat{F}=\hat{d}\)

A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velolcity of the 12kg mass is 4ms^-1. The kinetic energy of the other mass is

144 J
288 J
192 J
96 J

Solution

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate be fore coming to rest assuming that it faces constant resistance to motion ?

2.0 cm
3.0 cm
1.0 cm
1.5 cm

Solution

An automobile engine of mass M accelerates and a constant power pis applied by the engine.The instantaneous speed of the engine will be

[Pt/M]^1/2
[2Pt/M]^1/2
[Pt/2M]^1/2
[Pt/4M]^1/2

Solution

The potential energy of a conservative system is given by U =ay²– by, where y represents the position of the particle and a as well as b are constants.What is the force acting on the system ?

–ay
– by
2ay – b
b – 2ay

Solution

F = - ^dU⁄_dy = b - 2ay

A ball is allowed to fall from a height of 10 m. If there is 40% loss of energy due to impact, then after one impact ball will go up to

10 m
8 m
4 m
6 m

Solution

Kinetic energy of ball when reaching the ground

= mgh = mg × 10

Kinetic energy after the impact

=⁶⁰⁄₁₀₀ × mg × 10 = 6mg

If the ball rises to a height h, then mgh = 6 mg.

Hence, h = 6 m.

A body starts from rest and acquires a velocity V in time T.The work done on the body in time t will be proportional to

^V⁄_Tt
^V²⁄_Tt²
^V²⁄_T²t
^V²⁄_T²t²

Solution

Hail storms are observed to strike the surface of the frozen lake at 30° with the vertical and rebound at 60° with the vertical. Assume contact to be smooth, the coefficient of restitution is

e = ¹⁄_√3
e = ¹⁄₃
e = √3
e = 3

Solution

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete