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**Page No. 118****Q.1.** **Which of the following has more inertia:****(a) a rubber ball and a stone of the same size?****(b) a bicycle and a train?****(c) a five rupees coin and one-rupee coin?****Ans.** (a) Out of a rubber ball and a stone of same size, the mass of stone is larger; so stone has more inertia than a rubber ball.

(b) Out of a bicycle and a train, train has much larger mass than bicycle, so train has more inertia.

(c) Out of five rupees and one rupee coins, live rupee coin has larger mass and hence more inertia.**Q.2.** **In the following example, try to identify the number of times the velocity of the ball changes.****"A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team: Also identify the agent supplying the force in each case.****Ans. **The velocity of the ball changes four times in the following ways:

(a) Player 1 changes the velocity by kicking it.

(b) Player 2 changes the velocity by kicking it to the goal.

(c) Goalkeeper changes the velocity by, collecting the ball.

(d) Goalkeeper changes the velocity by kicking the ball.**Q.3.*** ***Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.****Ans. **The leaves of the tree have inertia of rest. When the branch is shaken it moves, but the leaves tend to be in a state of rest due to their inertia. Thus, they get detached from the tree.**Q.4.*** ***Why do you fall in the forward direction when a moving bus apply brakes to stop and fall back when it accelerates from rest?****Ans. **In a running bus, our speed is equal to the speed of the bus. As a moving bus brakes to a stop, the lower part of our body being in contact with the bus comes to rest, but the upper part due to inertia of motion, remains in the state of motion. Hence, we fall in the forward direction. When the bus accelerates from rest, the feet come into motion while upper part of the body remains at rest due to inertia of rest; hence, we fall backwards.**Page No. 126****Q.1. If action is always equal to the reaction, explain how a horse can pull a cart.**

Thus, to make the cart move, the horse bends forward and pushes the ground backward with its feet. When the forward force to the backward push of the horse is greater than the opposing frictional forces between tyres of cart and ground, the cart moves.

m = 50 g = 0.05 kg

Initial velocity of rifle (U) and bullet (u) = 0.

Final velocity of bullet, v = 35 m s

Recoil velocity of rifle, V = ?

By law of conservation of momentum,

mu + MU = mv + MV

Since u = U = 0,

Recoil velocity of gun in backward direction = 0.43 m s

Mass, m

m

Initial velocity, u

Final velocity, v

By law of conservation of momentum,

⇒

Velocity of second object = 1.16 m s

This is due to inertia of motion. If body is initially moving with some velocity on a smooth surface, then it will continue to move with same velocity, though the net external force acting on the body is zero. For example, when we stop pedalling a moving bicycle, the bicycle begins to slow down and finally comes to rest. This is again because of the friction forces acting opposite to the direction of motion. The force of friction opposes the motion of the bicycle. If there were no unbalanced force of friction and no air resistance, a moving bicycle would go on moving forever.

Time (t) = 20 s

Distance covered (S) = 400 m

Acceleration (a) = ?

Mass of truck (m) 7 tonne = 7000 kg

Force on truck We know ; (F) = ?

We know ;

S

400 = 0 × 20 1/2 × a × (20)

400 = 200a

a =

Force on truck (F) = ma

= 7000 x 2

= 14000 N

Final velocity, v = 0;

distance travelled, s = 50 m

Acceleration,

Force exerted = ma

= 1 kg x (- 4) ms

Friction = 4 N against the direction of motion.

Force of friction exerted by the tracks = 5000 N

As the force of friction always acts opposite to the direction of applied force.

Net accelerating force of engine

= 40000 _ 5000 = 35000 N

(b) Mass 5 wagons

= 2000 × 5 = 10000 kg

We know : F = ma

35000 = 10000 × a

(c) Force of wagon 1 on wagon 2

F = ma

= 8000 × 3.5 = 28000 N

F = ma = 1500 kg x ( -1.7 ms

= -2,550 kg ms

= -2550 N

(Negative sign indicates that the force is in a direction opposite to the motion of the vehicle.)

Thus friction = - 200 N

Initial velocity of first object, u

Initial velocity of second object, u

Combined mass after collision = m

From the principle of conservation of momentum

⇒ 1.5 x 2.5 + 1.5 x (-2.5) = (3.0) v

⇒ 0 = 3.0 v

or v = 0 ms

That is combined object comes to rest after collision.

The truck is massive, so it has very large inertia. Moreover, it is parked on the ground; there is a frictional force between the truck and ground. The force exerted by us on the truck is insufficient to overcome the force of friction; so these are balanced forces on the truck; (force exerted by us plus force of friction in opposite direction); that is net force is zero; and hence, the truck does not move.

Initial velocity of ball (u

Final velocity of ball (u

(Negative sign denotes that ball is moving in opposite direction)

Initial momentum of ball = mu

Final momentum of ball = mu

Change in momentum = Final momentum + Initial momentum

= 1 2 = 3 Ns

Negative sign denotes that change in momentum is in the direction opposite to the direction of initial momentum of the ball.

Initial velocity of bullet (u) =

Final velocity of bullet (V) = 0

Time (t)

Acceleration on bullet (a) = ?

Force acting on wooden block (F) = ?

Distance penetrated by bullet (s) = ?

We know;

v = u at

0 = 150 a ×

a × 0.03 = 150

We know; S = ut at

= 150 × 0.03 × (5000) × (0.03)

We know, F = ma

Force acting on bullet (F)

= 0.01 × ( 5000) = 50N

Negative sign denotes that wooden block exerts force in the direction, opposite to the direction of motion of the bullet.

u

For wooden block :

m

u

Momentum just before collision

= m

= 1 × 10 5 × 0

= 10 kg ms

Mass after collision = (m

= I 5 = 6 kg

Let velocity after collision = v

Momentum after collision = 6 × v

Using the law of conservation of momentum

Momentum after collision = Momentum before collision

6 × v =10

v = = 1.67 ms

Initial velocity (u) = 5 ms

Final velocity (v) = 8 ms

Time (t) = 6 s

Initial momentum (P

Final momentum (P

Force exerted on the object (F) =

Mass, m = 10 kg ; Initial velocity, u = 0,

Acceleration, a= 10 ms

Final velocity, v =

Momentum transferred = mv = 10 kg x 4 ms

= 10 kg x 4 ms

= 40 kgms

Time (second) | Distance (metre) |

0 | 0 |

1 | 1 |

2 | 8 |

3 | 27 |

4 | 64 |

5 | 125 |

6 | 216 |

7 | 243 |

**(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing or zero?****(b) What do you infer about the forces acting on the object?****Ans.** (a) Initial speed, u = 0

Since s =

Thus, acceleration is increasing with time.

(b) Force, F = ma ⇒ F ∝ a

Since acceleration is increasing with time, force is also increasing with time.**Q.2.** **Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms ^{2}. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)**

Let each person applies force F on the motor car. Thus, the force ‘2F applied by 2 persons balances the frictional fore due to ground force applied by three persons = 3F

Effective force = 3F - friction = 3F - 2 F = F

Thus force F produces an acceleration of 0.2 ms

F = ma = 1200 x 0.2 = 240 N

Thus, force applied by each person = 240 N

Initial velocity of hammer, u = 50 ms

Final velocity of hammer, v = 0

Time, t = 0.01 s

Acceleration,

= - 5000 ms

Force applied by nail, F = ma

= (0.5) x (-5000 ms

= - 2500 N (opposite to motion of hammer)

Initial velocity of car, u = 90 km/h

Final velocity of car, v = 18 km/h

Time, t = 4s

Acceleration,

Momentum change = mv - mu

= 1200 (5 - 25) = - 24000 kg ms

Magnitude of force, F = ma

= 1200 x (- 5) = - 6000 N

Acceleration, momentum change and force are opposing the motion of motorcar, as indicated by negative sign.

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