Laws of Motion

• Question 573

## A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms'.The kinetic energy of the other mass is256 J 486 J 524 J 324 J

Solution

B.

486 J

The linear momentum of the exploding part will remain conserved.

Applying coservation of linear momentum,

m1 u1 =m2 u2

Here, m1 = 18kg, m2 = 12kg

u1 = 6ms-1, u2 = ?

Thus kinetic energy of 12 kg mass

Question 574

## A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle θ. The frictional forceconverts translational energy to rotational energy dissipates energy as heat decreases the rotational motion decreases the rotational and translational motion

Solution

A.

converts translational energy to rotational energy

When a body rolls down without slipping along an inclined plane of inclination θ, it rotates about a horizontal axis through its centre of mass and also its centre of mass moves. Therefore, rolling motion may be regarded as a rotational motion about an axis through its centre of mass plus a translational motion of the centre of mass. As it rolls down, it suffers loss in gravitational potential energy provided translational energy due to frictional force is converted into rotational energy.

Question 575

## A block B is pushed momentarily along a horizontal surface with an initial velocity v. If µ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time$\frac{\mathrm{\nu }}{\mathrm{g\mu }}$ $\frac{\mathrm{g\mu }}{\mathrm{\nu }}$ $\frac{\mathrm{g}}{\mathrm{\nu }}$ $\frac{\mathrm{\nu }}{\mathrm{g}}$

Solution

A.

$\frac{\mathrm{\nu }}{\mathrm{g\mu }}$

Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,
force applied  =  frictional force

μmg = m$\mathrm{\alpha }$

where μ - constant of proportionality

Question 576

## Diwali rocket is ejecting 50 g of gases/s at a velocity of 400 m/s. The accelerating force on the rocket will be22 dyne 20 N 20 dyne 100 N

Solution

B.

20 N

The accelerating force on the rocket

= upward thrust = $\frac{∆\mathrm{m}}{∆\mathrm{t}}.\mathrm{u}$

Given:-

,

ν  = 400 m/s

Accelerating force = 50 × 10 ×  400

= 20 N