## JAC Board Class 9th Science Solutions Chapter 9 Force and Laws of Motion

**JAC Class 9th Science Force and Laws of Motion InText Questions and Answers**

Page 118

Question 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 a one – rupee coin?

Answer:

(a) A stone

(b) A train

(c) A five – rupee coin

As the mass of an object is a measure of its inertia, objects with more mass have more inertia.

Question 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.

Answer:

The velocity of ball changes four times.

Agent supplying the force | Change in velocity of ball |

(a) First player kicks a football. | (a) Velocity changes from ‘zero’ to ‘u’. |

(b) Second player kicks the football towards the goal. | (b) Velocity changes again by change in direction. |

(c) The goalkeeper collects the football. | (c) Velocity becomes zero. |

(d) Goalkeeper kicks it towards a player of his team. | (d) Change in velocity takes place. |

Question 3.

Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer:

Before shaking the branches, leaves are at rest. When branches are shaken, they come in motion while the leaves tend to remain at rest due to inertia of rest. As a result, leaves get detached from the branches and fall down.

Question 4.

Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer:

1. When a moving bus brakes to a stop: When the bus is moving, our body is also in motion. But due to sudden brakes, the lower part of our body comes to rest as soon as the bus stops. But the upper part of our body continues to be in motion and hence we fall in forward direction due to inertia of motion.

2. When the bus accelerates from rest: When the bus is stationary, our body is at rest but when the bus accelerates, the lower part of our body, being in contact with the floor of the bus, comes in motion but the upper part of our body remains at rest due to inertia of rest. Hence we fall in backward direction.

Page 126

Question 1.

If action is always equal to the reaction, explain how a horse can pull a cart.

Answer:

The horse pulls the cart with a force (action) in the forward direction. The cart also pulls the horse with an equal force (reaction) in the backward direction. The two forces get balanced While pulling the cart, the horse also pushes the ground with its feet in the backward direction, the reaction of the earth makes it move in the forward direction along with the cart.

Question 2.

Explain, why it is difficult for a fireman to hold a hose, which ejects a large amount of water at a high velocity.

Answer:

The water that is ejected out from the hose in the forward direction, comes out with a large momentum and an equal amount of momentum is developed in the hose in the opposite direction and hence the hose is pushed backward It hence becomes difficult for a fireman to hold a hose which experiences this large momentum.

Question 3.

From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35m/s. Calculate the initial recoil velocity of the rifle.

Answer:

(m_{1}) Mass of rifle = 4kg

(m_{2}) Mass of bullet = 50g = 0.05kg

(v_{2}) Velocity of bullet = 35m/s

(v_{1}) Recoil velocity of rifle = v_{1}

According to the law of conservation of momentum,

momentum of rifle = momentum of bullet

m_{1}v_{1} = v_{2}m_{2}

4 kg × v_{1} = 0.05 × 35m/s

v_{1 }= \(\frac{0.05 \times 35}{4}\) = \(\frac{1.75}{4}\) = 0.4375 m/s.

Question 4.

Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2m/s and 1m/s respectively. They collide and after the collision, the first object moves at a velocity of 1.67m/s. Determine the velocity of the second object.

Answer:

m_{1} = 100g = 0.1kg

m_{2}= 200g = 0.2kg

u_{1} = 2m/s

u_{2} = 1m/s

After collision

v_{1} = 1.67m/s

v2=?

m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}

(0.1 × 2) + (0.2 × 1)

= (0.1 × 1.67) + (0.2 × v_{2})

0. 2 + 0.2 = 0.167 + 0.2 v_{2}

0. 4 = 0.167 + 0.2 v_{2}

\(\frac{0.4-0.167}{0.2}\) = v_{2}

\(\frac{0.233}{0.2}\) = 1.165 m/s

The velocity of the second object is 1.165 m/s.

**JAC Class 9th Science Force and Laws of Motion Textbook Questions and Answers**

Question 1.

An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non – zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer:

When an object experiences a net zero external unbalanced force, in accordance with the second law of motion, its acceleration is zero. If the object was initially in a state of motion, then in accordance with the first law of motion, the object will continue to move in the same direction with the same speed It means that the object may be travelling with a non – zero velocity but the magnitude as well as direction of velocity must remain unchanged or constant throughout.

Question 2.

When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:

The carpet with dust is in state of rest. When it is beaten with a stick the carpet is set in motion, but the dust particles remain at rest. Due to inertia of rest, the dust particles retain their position of rest and fall down due to gravity.

Question 3.

Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:

Owing to sudden jerks or due to the bus taking sharp turn on the road, the luggage may fall down from the roof because of its tendency to continue moving in the original direction. To avoid this, the luggage is tied with a rope on the roof.

Question 4.

A batsman hits a cricket ball which then rolls on a level ground After covering a short distance, the ball comes to rest. The ball slows to a stop because

(a) the batsman did not hit the ball hard enough.

(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion.

(d) there is no unbalanced force on the ball, so the ball would want to come to rest.

Answer:

(c) there is a force on the ball opposing the motion.

Question 5.

A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes.

(Hint: 1 tonne = 1000 kg).

Answer:

u = 0m/s

m = 7 tonnes = 7 × 1000kg = 7000kg

s = 400m, t = 20s,

a = ? F = ?

s = ut +\(\frac{1}{2}\) at^{2}

400 = (0 × 20) + \(\frac{1}{2}\) a (20)^{2}

⇒ \(\frac{400 \times 2}{(20)^{2}}\) = a

or, a = 2m/s^{2}

Force (F) = ma = 7000 × 2 = 14000N

Question 6.

A stone of 1kg is thrown with a velocity of 20 ms^{-1 }across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice?

Answer:

m = 1kg, u = 20m/s,

s = 50m, v = 0

F = ? a = ?

v^{2} – u^{2} = 2as

(0)^{2} – (20)^{2} = 2a (50)

-400 = 100a

a = \(\vec{a}\) = ( – 4m/s^{2}) = – 4N

Question 7.

A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate.

(a) the net accelerating force and

(b) the acceleration of the train.

Answer:

(a) The net accelerating force = Force exerted by the engine – friction force = 40000 N – 5000 N = 35000 N

(b) The acceleration of the train (a) = ?

F = 35000 N

Mass of 5 wagons pulled by the engine = 5 × 2000 = 10000kg

F = ma

35000 = 10000 × a

a = 35000/10000 = 3.5 m/s^{2}

Question 8.

An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7ms^{2}?

Answer:

Mass = 1500kg

a = -1.7 m/s^{2}

F = ?

F = m × a = 1500 × (- 1.7) = – 2550 N

The force between the vehicle and road is – 2550 N.

Question 9.

What is the momentum of an object of mass m, moving with a velocity v?

Answer:

(a) (mv)^{2}

(b) v^{2}

(c) \(\frac{1}{2}\)mv^{2}

(d) mv

Question 10.

Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer:

The cabinet will move with constant velocity only when the net force on it is zero. Force of friction on the cabinet is 200 N in a direction opposite to the direction of motion of the cabinet.

Question 11.

Two objects, each of mass 1.5kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5ms^{1} before the collision during which they stick together. What will be the velocity of the combined object after collision?

Answer:

Mass of the objects, m_{1} = m_{2} = 1.5kg

Velocity of first object, v_{1} = 2.5 m/s

Velocity of second object, v_{2} = – 2.5 m/s

Momentum before collision = m_{1}v_{1} + m_{2}v_{2}

= (1.5 × 2.5)+ (1.5 × (- 2.5)) = 0

Mass of combined object = m_{1} + m_{2} = 1.5 + 1.5 = 3.0kg

After collision, v = ?

According to the law of conservation of momentum,

Momentum after collision = Momentum before collision

mv = 0

or v = 0 ms^{-1}

(since mass cannot be zero)

Question 12.

According to the third law of motion when we push an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Answer:

Action and reaction always act on different bodies, so they cannot cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.

Question 13.

A hockey ball of mass 200 g travelling at 10 ms^{-1} is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^{-1} Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer:

Mass of ball, m = 200g = 0.2kg

Initial speed of ball, u = 10m/s

Final speed of ball, v = – 5 m/s

Initial momentum of the ball

= mu = 0.2kg × 10m/s = 2kg m/s

Final momentum of the ball

= mv = 0.2kg × (- 5 m/s) = – 1kg m/s

Hence, change in momentum

= difference in the momentum = (-1) – 2 = -3kg m/s

Question 14.

A bullet of mass 10g travelling horizontally with a velocity of 150ms^{-1} strikes a stationary wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

m = 10g = 0.01kg, u = 150ms^{-1},

v = 0, t = 0.03s

a = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) = \(\frac{0-150}{0.03}\) = – 5000ms^{-2},

The distance of penetration of the bullet into the block,

s = ut +\(\frac{1}{2}\) at^{2}

= (150 × 0.03) + \(\frac{1}{2}\) × (- 5000) × (0.03)^{2}

= 4.5 – 2.25 = 2.25 m

The magnitude of the force exerted by the wooden block on the bullet

= ma = 0.01 × 5000 = 50N.

Question 15.

An object of mass 1kg travelling in a straight line with a velocity of 10ms^{-1} collides with and sticks to a stationary wooden block of mass 5kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Answer:

m_{1}= 1 kg

v_{1} = 10 m/s

Mass of wooden block = 5 kg

m_{2} = 5kg + 1kg (combined object)

= 6kg

Velocity of combined object = v_{2 }= ?

Momentum before impact (p_{1}) = m_{1}v_{1} = 1 × 10 = 10 kg m/s

∴ Momentum before impact = Momentum after impact

m_{1}v_{1} = m_{2}v_{2}

10Kg m/s = 6v_{2}

\(\frac{10}{6}\) = v_{2 }or v_{2} = 1.76 m/s

Question 16.

An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms^{-1} to 8 m s^{-1} in 6s. Calculate the initial and final momentum of the object. Also, find out the magnitude of the force exerted on the object.

Answer:

Given:

m = 100 kg,

u = 5 m/s,

v = 8 m/s,

t = 6s

p_{1} = ?

P_{2} = ?

F = ?

Initial momentum, p_{1} = mu = 100 × 5 = 500 kg m/s

Final momentum, p_{1} = mv = 100 × 8 = 800 kg m/s

Force exerted on the object, F = ma

= 100 \(\left(\frac{v-u}{t}\right)\) = 100 \(\left(\frac{8-5}{6}\right)\)

= 100 × \(\frac{3}{6}\) = 50 N

Question 17.

Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect and as a result, the insect died Rahui while putting on entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Answer:

Rahul gave the correct explanation that both the motorcar and the insect experienced the same force and a change in their momentum. As per the law of conservation of momentum, when two bodies collide:

Initial momentum before collision = Final momentum after collision

m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}

Equal force is exerted on both the bodies but because the mass of insect is very small, it will suffer a greater change in velocity.

Question 18.

How much momentum will a dumb bell of mass 10kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10ms^{2}.

Answer:

Here, m = 10kg, u = 0,

s = 80 cm = 0.80 m, a = 10 m/s^{-2}

Let v be the velocity gained by the dumb bell as it reaches the floor.

As v^{2} – u^{2} = 2as

v^{2} – 0^{2} = 2 × 10 × 0.80 = 16

v = 4 ms^{-1}

Momentum transferred by the dumb bell to the floor:

p = mv = 10 × 4 = 40kg ms^{-1}

Page 130

Question A1.

The following is the distance – time table of an object in motion:

Time in seconds | Distance in metres |

0 | 0 |

1 | 1 |

2 | 8 |

3 | 27 |

4 | 64 |

5 | 125 |

6 | 216 |

7 | 343 |

(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?

Answer:

Time (s) | Distance (m) |
Vloctiy, v = \(\frac{S_{2}-S_{1}}{t_{1}-t_{1}}\left(\mathrm{~s}^{-1}\right)\) | Acceleration, a = \(\frac{v_{2}-v_{1}}{t_{2}-t_{1}}\left(m^{-2}\right)\) |

0 | 0 | 0 | 0 |

1 | 1 | \(\frac{1-0}{1-0}=1 \mathrm{~ms}^{-1}\) | \(\frac{1-0}{1-0}=1 \mathrm{~ms}^{-2}\) |

2 | 18 | \(\frac{8-1}{2-1}=7 \mathrm{~ms}^{-1}\) | \(\frac{7-1}{2-1}=6 \mathrm{~ms}^{-2}\) |

3 | 27 | \(\frac{27-8}{3-2}=19 \mathrm{~ms}^{-1}\) | \(\frac{19-7}{3-2}=12 \mathrm{~ms}^{-2}\) |

4 | 64 | \(\frac{64-27}{4-3}=37 \mathrm{~ms}^{-1}\) | \(\frac{37-19}{4-3}=18 \mathrm{~ms}^{-2}\) |

5 | 125 | \(\frac{125-64}{5-4}=61 \mathrm{~ms}^{-1}\) | \(\frac{61-37}{5-4}=24 \mathrm{~ms}^{-2}\) |

6 | 216 | \(\frac{216-125}{6-5}=91 \mathrm{~ms}^{-1}\) | \(\frac{91-61}{6-5}=30 \mathrm{~ms}^{-2}\) |

7 | 343 | \(\frac{343-216}{7-6}=127 \mathrm{~ms}^{-1}\) | \(\frac{127-91}{7-5}=36 \mathrm{~ms}^{-2}\) |

(a) There is an unequal change of distance covered in equal intervals of time. Thus, the given object is having a non – uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.

(b) The object is in an accelerated condition. According to Newton’s second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. So, we can say that unbalanced forces are acting on the object.

Question A2.

Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level roa(d) The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s 2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort).

Answer:

Mass of the motorcar = 1200kg.

Only two persons manage to push the car with uniform velocity. Hence, the acceleration acquired by the car is given by the third person alone.

Acceleration produced by the car when it is pushed by the third person is, ‘a’

= 0.2 m/s^{2}.

Let the force applied by the third person be F.

From Newton’s second law of motion, Force = Mass × Acceleration => F = 1200 × 0.2 = 240 N.

Thus, the third person applies a force of magnitude 240 N. Hence, each person applies a force of 240 N to push the motorcar.

Question A3.

A hammer of mass 500 g, moving at 50ms^{1}, strikes a nail. The nail stops the hammer in a very short time of 0.01s. What is the force of the nail on the hammer?

Answer:

Mass of the hammer,

m = 500g = 0.5kg.

Initial velocity of the hammer,

u = 50m/s.

Time taken by the nail to stop the hammer, t = 0.01s.

Velocity of the hammer, v = O (since the hammer finally comes to rest). From

Newton’s second law of motion,

Force, F = \(\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\) = \(\frac{0.5(0-50)}{0.01}\)

= – 2500N

The hammer strikes the nail with a force of -2500N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e.. + 2500 N.

Question A4.

A motorcar of mass 1200kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18km/h in 4s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Answer:

Mass of the motorcar, m = 1200kg.

Initial velocity of the motorcar, u = 90km/h = 25m/s.

Final velocity of the motorcar,

v = 18 km/h = 5m/s.

Time taken, t = 4s.

According to the first equation of motion:

v = u + at

=> 5 = 25 + a(4) or, a = -5m/s^{2}

Negative sign indicates that it is a retarding motion, i.e., velocity is decreasing.

Change in momentum = mv – mu

= m(v – u)= 1200 (5 – 25)

= – 24000 kg ms^{-1}

Force = Mass × Acceleration

= 1200 × (- 5) = – 6000 N