# Grade 9 – Physics – Force and Laws of Motion – Important questions and answers

Question 1: State the law of conservation of momentum.
Answer: If the external force acting on a system of bodies is zero, the net momentum of the system remains conserved.

Question 2:What is the name given to the product of mass and velocity of a body?

Question 3: A body of mass “m” is acted upon by a force ‘F’ which produces an acceleration ‘a’ in it. If the force is halved, how will the acceleration change?

Answer: When force becomes F/2, the acceleration reduces to a/2 because a $\propto$ F.

Question 4: State the relation between force and momentum.

Answer: Force $=$ Rate of change of momentum or ${F}=\frac{p_{2}-p_{1}}{t}$

Question 5:Two friends on roller-skates are standing 5 m apart facing each other. One of them throws a ball of 2 kg towards the other who catches it. How will this activity affect the position of the two? Explain your answer.

Answer: Separation between them will increase. Initially the momentum of both of them is zero as they are at rest. In order to conserve the momentum the one who throws the ball would move backward. The second will experience a net force after catching the ball and therefore will move backwards that is in the direction of force.

Question 6: Why does a boxer move his hand backwards to minimise the effect of oncoming punch?

Answer: A passenger travelling in a bus shares the motion of the bus. Thus, the passenger’s body is moving in the direction of the bus with the same speed. When a passenger jumps out of a fast moving bus, his feet on touching the ground comes to rest, whereas the upper part of his body continues to move forward due to inertia of motion. As a result, if he does not run forward, he falls with his face downwards.

Question 7: Explain the following :
(i) Why it is difficult to walk on marshy land?
(ii) Why does an air filled balloon rise up slightly when punctured from below?
(iii) Why does a swimmer push water backward with his hands in order to swim in forward direction?

Answer: (i) It is difficult to walk on marshy land because when we push the ground with our feet, the ground yields and gives no reaction.
(ii) As the air comes out of the punctured balloon it applies an upward reaction on the balloon. Thereby making it rise up slowly.
(iii) According to Newton’s third law of motion, when the swimmer pushes the water backwards the water applies a forward reaction on him. This makes the swimmer move forward.

Question 8: When a ball falls on the earth, the earth also moves up to meet it. But the motion of the earth is not noticeable. Explain why?

The acceleration produced in the earth is given by $a= \frac{F}{M_e}$ , $M_e$ is the mass of earth.

Since mass of earth is very large, so acceleration produced in it is negligible. Hence the motion of the earth is not noticeable. However, the acceleration produced in the ball is given by $a= \frac{F}{m}$.

Since m is very small, so acceleration produced in the ball is large. Hence the downward motion of the ball is easily noticeable.

Question 9:Find an expression for the recoil velocity of a gun, when it fires a bullet.

Let a gun has a mass $\mathrm{M}$ and a bullet of mass $m$ is fitted into it. Initially, gun and bullet both are at rest and their total linear momentum is zero.

When the gun is fired, the bullet immediately comes out and moves forwards with a muzzle velocity v. Let the gun gains a velocity $\mathrm{V}$ in the process, then Final momentum of bullet $=m v$
and $\quad$ Final momentum of gun $=\mathrm{MV}$
$\therefore \quad$ Total momentum of gun-bullet systems $=\mathrm{MV}+m v$ According to law of conservation of momentum
Total final momentum $=$ total initial momentum
$\therefore \mathrm{MV}+m v=0$
$\Rightarrow \mathrm{V}=-\frac{m v}{\mathrm{M}}$
The $-$ve sign of $\mathrm{V}$ shows that the gun recoils in a direction opposite to the motion of bullet. Moreover, recoil velocity of gun is inversely proportional to its mass.

Question 10: Indicate the forces of action and reaction in the following cases :
(i) a man standing on the ground.
(ii) a stone suspended by a thread from the ceiling.
(iii) a book lying on the table.
(iv) motion of moon around the earth, and
(v) when a magnet attracts a piece of iron.

Answer. (i) The man exerts a downward force equal to his weight on the ground (action). The ground exerts an equal upward force on the man (reaction).
(ii) The weight of the stone acts vertically downwards (action). A tension equal to the weight of the stone acts vertically towards the thread.
(iii) The book exerts a downward force equal to its weight on the table (action). The table exerts an equal force on the book in the upward direction (reaction).
(iv) The moon attracts the earth with some force (action). The earth also attracts the moon with an equal and opposite force (reaction).
(v) When a magnet and a piece of iron are brought close to each other, the magnet attracts the iron piece with a certain force (action). The iron piece also attracts the magnet with an equal and the opposite force (reaction).

Question 11: Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown below. Calculate the acceleration and frictional force of the floor on the ball.

From the graph, we find that,

$u=80 \ ms^{-1}, v=0, t=8 s$

From$\begin{array}{l}v=u+a t \\0=80+a \times 8 \\\mathrm{or} , \ a=-\frac{80}{8}=-10 \ ms^{-1}\end{array}$

$m=50 \ \mathrm{g}=\frac{50}{1000} \ \mathrm{kg}=\frac{1}{20} \ \mathrm{kg}$

$\therefore\mathrm{F}=m a=\frac{1}{20}(-10)=-0.5 \ \mathrm{N}$

Question 12: Calculate the force required to stop a car of mass 1000 kg and a loaded truck of mass 10,000 kg in 2 seconds if they are moving with the same velocity of $5 ms^{- 1}$ .

Car:

\begin{aligned}m &=1000 \ \mathrm{kg}, u=5 \ \mathrm{ms}^{-1}, t=25, v=0 \\v &=u+a t \\\Rightarrow v &=5+a \times 2 \\\Rightarrow a &=-\frac{5}{2}=-2.5 \ \mathrm{ms}^{-2} \\\text {Retarding force } &=m a=1000 \times 2.5=2500 \ \mathrm{N} \end{aligned}

Truck:

\begin{aligned}u &=5 \ \mathrm{ms}^{-1}, t=2 \ \mathrm{s}, v=0 \\\therefore a &=-2.5 \ \mathrm{ms}^{-1}\end{aligned}

$\\\text{Retarding force} =ma=10,000 \times 2.5=25000 \ \mathrm{N}$

Question 13: Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attain the speed v again before striking the ground. It implies that the magnitude of initial and final momentum of the ball are same. Yet it is not an example of conservation of momentum. Explain Why?

Answer: Momentum of a system remains conserved if no external force acts on the system. In the given example, there is gravitational force acting on the ball, so it is not an example of conservation of momentum.

Question 14: A car with a dead battery is to be pushed for some time, so as to start it. Why? What does this example signify?

Answer: If we want to start a car with a dead battery, then we have to push it along a straight road to give it a certain minimum speed (of about 1 $ms^{- 1}$ ). When one or two persons give it a sudden push, they are unable to give desired speed to the car and it does not start. However,a continuous push over sometime results in a gradual acceleration of the car to this maximum speed and car starts.
This example signifies that effect of a force depends on :

(i) magnitude of the force and
(ii) time for which force is applied.

Question 15:

(i) What do you understand by the term force?
(ii) State the law of conservation of momentum.
(iii) Define newton.
(iv) What do you understand by unbalanced force ?
(v) Define the term impulse.

Answer. (i) Force: A force is an external physical agent which changes or tends to change the state of rest or of uniform motion, direction of motion, or the shape or size of the body.
(ii) Conservation of momentum : If external force acting on a system of bodies is zero, the net momentum of the system remains conserved.
(iii) One newton is the force acting on a body of mass one kg produces an acceleration of $1\ ms^{-1}$.
(iv) If a number of forces are acting on a body such that resultant of the forces acting on a body is not zero, then such a force is called an unbalanced force.
(v) When a larger force is applied on a body for a very short interval of time then product of force and time is called impulse.

Question 16: Why does a boat tend to leave the shore, when passengers are alighting from it?

Answer. When the passengers alight from the boat, they push the boat in backward direction. As a result, the boat has a tendency to slip back into water. This difficulty is usually overcome by the boatman by tying the boat to some rigid support.

Question 17:

A bus of mass 5000 kg starts from rest and rolls down a hill. If it travels a distance of 200 m in 10 s, calculate
(i) acceleration of the bus and
(ii) the force acting on the bus

\begin{aligned}m &=5000 \mathrm{kg}, u=0, s=200 m, t=10 \ \mathrm{s} \\a &=?, F=? \\\mathrm{s} &=u t+\frac{1}{2} a t^{2} \\ 200 &=0+\frac{1}{2} a \times(10)^{2}\end{aligned}
$\therefore$ Acceleration of the bus force acting on the bus,
$a=4 \ m s^{-2}$
$\mathrm{F}=m a$

F = 5000 x 4 = 20,000 kg $ms^{- 2}$
F = 20,000 N

Question 18 : If action is always equal to reaction, Explain how a horse can pull a cart ?

Answer: The horse pushes the ground in the backward direction. The ground exerts a reaction force on horse and cart system to push them forward. When the reaction force exceeds the force of friction between the wheels of cart and the ground, the cart is pushed forward.

Question 19: Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2m/s and 1 m/s respectively. They collide and after collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.

Before collision,
$\begin{array}{l} m_{1}=100 g=\frac{100}{1000}=0.1 \ \kg \\ m_{2}=200 \ {g}=\frac{200}{1000}=0.2 \ {kg} \\ u_{1}=2 \ {ms}^{-1}, \ u_{2}=1 \ {ms}^{-1} \end{array}$

After collision ,$\begin{array}{l}m_{1}=100 \ {g}=0.1 \ {kg}, m_{2}=200 \ {g}=0.2 \ {kg} \\ v_{1}=1.67 \ {ms}^{-1}, v_{2}=? \end{array}$

Let velocity of second object after collision be $v_{2} \ {ms}^{-1} .$ since there is no external force on the system, So total momentum before collision $=$ total momentum after collision
or
$\begin{array}{l}{m}_{1} {u}_{1}+{m}_{2} {u}_{2} ={m}_{1} v_{1}+{m}_{2} v_{2} \\0.1 \times 2+0.2 \times 1 =0.1 \times 1.67+0.2 \times v_{2} \\0.2+0.2 =0.167+0.2 v_{2} \\0.2 v_{2} =0.4-0.167=0.233 \\v_{2} =\frac{0.233}{0.2}=1.165 \ {ms}^{-1}\end{array}$

Question 20: Two balls of the same size but of different materials, rubber and iron, are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer.

Answer : Yes both the balls will start rolling due to inertia of motion. The train stops due to brakes. But the balls continue to move in the direction of the train. The two balls will not roll with the same speed. In fact, the iron ball being heavier than the rubber ball shall move at a lower speed.

Question 21: Why does a bullet fired against a glass window pane make a hole in it, and the glass pane does not crack, while a stone striking the same glass pane will smash it?

When a bullet strikes the glass pane, the part of the glass pane which comes in contact with the bullet immediately shares the large velocity of the bullet and flies away making
a hole. The remaining part of the glass, due to inertia of rest, remains at rest and is not smashed.
However when a slow-moving stone strikes the glass pane, a much larger area of the glass pane gets enough time to share the velocity of the stone. As a result, the glass cracks in all directions.

Question 22: State the meaning of balanced forces.
Answer. If the resultant of all the forces acting on a body is zero, the forces are called balanced forces.

Question 23: Name the force which arises between two surfaces in contact.

Question 24: State Newton’s third law of motion.
Answer. According to Newton’s third law of motion, whenever one body exerts a force on another body, the second body exerts an equal and opposite force on the body.

Question 25:

A stone of size 1 kg 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 50 m. What is the force of friction between the stone and the ice?

$\begin{array}{ll} m=1 \ {kg}, & u=20 \ {ms}^{-1} \\ v=0, & \ {s}=50 \ {m} \end{array}$
( i) $v^{2}-u^{2}=2 a s,$ we get
$a=\frac{v^{2}-u^{2}}{2 s}=\frac{0-400}{2 \times 50}=-4 \ m s^{-2}$
(ii) $\quad \mathrm{F}=m a,$ we get
$\mathrm{F}=1 \times(-4)=-4 \ {N}$
$\therefore$ Force of friction $=4 \ {N}$