# Illustrations

A vehicle is travelling at a velocity of 10 m s^{–1}. An accelerating force of 240 N is applied for a duration of 15 s. The mass of the vehicle is 1200 kg. What is the final velocity of the vehicle?

*u* = 10 m s^{–1}, *F* = 240 N, *t* = 15 s, *m* = 1200 kg, *a* = ?, *v* = ?

*F* = *ma* or

*v* = *u* + *at*

= 10 m s^{–1} + (0.2 m s^{–1} 15 s)

= 10 + 3 = 13 m s^{–1}

A gun having a mass of 5000 g fires a 20 g bullet at a speed of 500 m s^{–1}. Find the recoil velocity of the gun.

Mass of the bullet, *m*_{1} = 20 g = 0.02 kg

Velocity of the bullet, *v*_{1} = 500 m s^{–1}

Mass of the gun, *m*_{2} = 5000 g = 5 kg

Let the recoil velocity of gun be *v*_{2 }m s^{–1}.

According to the law of conservation of momentum,

Momentum of bullet = Momentum of gun

*m*_{1} *v*_{1} = *m*_{2} *v*_{2}

0.02 500 = 5 *v*_{2}

A book of mass 0.5 kg lies on the surface of a table. The coefficient of friction between the book and the surface of the table is 0.6. Calculate the force required to just move it (take *g* = 10 m s^{–2}).

Another identical book is kept on the book on the table. How is the force needed to move it affected?

Given *m* = 0.5 kg,

*R* = *W* = *mg* = 0.5 × 10 = 5N, μ = 0.6

In the limiting condition,

Force required = Force of friction

∴ Force required *F* = *f* = μ*R* = 0.6 5=3Ν

When another book is placed over the first book, the reaction becomes

*R*´ = *W*´ = 2*W*

= 2 0.5 10 = 10 N

Force required *F*´ = *f*´ = μ*R*´

= 0.6 10=6Ν

A block of weight 10 N lies on a rough table. It just moves when a force of 7 N is applied on it. Calculate the coefficient of limiting friction between the block and the table.

In the limiting condition, force of friction *f* = applied force *F* = 7 N,

Normal reaction, *R* = weight *W* = 10 N

From the relation *f* = μ*R*,

Coefficient of friction

Two bodies each of mass 25 kg are separated by a distance 1 m. Find the force between them.

What will be the force, if the distance between the masses is doubled?

[Assume G = 6.673 × 10^{−11} N m^{2} kg^{–2}].

Mass of the first body = *m*_{1} = 25 kg

Mass of the second body = *m*_{2} = 25 kg

Separation between them = *r*_{1} = 1 m

We know that, the gravitational force

*F*_{1} = 4.17 × 10^{−8} Ν

For the second part *r*_{2} = 2*r*_{1}

*F*_{2} = 1.04 × 10^{−8} Ν

Calculate the values of the acceleration due to gravity (i) at a depth of 8 km and (ii) at an altitude of 32 km from the surface, given that the value of *g* on the surface at sea level is 9.8 m s^{–2}.

(i) The value of *g* at any depth is given by

= 9.788 m s^{–2}

(ii) The value of *g* at an altitude of 32 km is

*g* = 9.70 m s^{–2}