# SSC Worksheet for chapter-3 Laws Of Motion class 9

## Fill in the Blanks

1. Tendency of a body to resist ____________ is called inertia
2. Newton is the SI unit of ________
3. If unbalanced forces acting on a body is halfed, acceleration with which it moves becomes______ of its initial acceleration
4. _______ forces do not change the state of rest or of motion of an object.
5. Action and reaction forces do not cancel out because they act on _______ bodies
6. Principle used for motion of a rocket is based on Newton’s _________law of motion
7. State with suitable illustrations the three laws of motion of Newton.
8. State with examples the three types of inertia
9. State principle of conservation of momentum. Derive the principle based on Newton’s third law of motion.
10. Write a note on momentum.
11. The velocity of body of mass 1 kg changes from 15m/s to 25m/s in 5 seconds, due to the action of a constant force acting on it. Find magnitude of the force
12. A shell of mass 6 kg is fired from a gun of mass 600 kg. The recoil velocity of the gun is 3m/s; find the muzzle velocity of the shell.
13. A particle of mass 20kg, which is initially at rest, is acted upon by a constant force of 10 N for 10 seconds. Find velocity attained by the particle after 10s and distanced covered by it in the same time.
14. A billiard ball of mass 1 kg, moving with a velocity of 4 m/s collides with another billiard ball of mass 0.5 kg moving with a velocity of 2 m/s in the same direction as the first ball. After collision, the first ball attains a velocity of 8/3 m/s. Find the velocity of the second ball after collision.
15. A bullet of mass 10 g moving with a speed of 1.5 m/s gets embedded in a stationary wooden block of mass 90 g. Find the speed attained by the block containing the embedded bullet

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### Solutions

1. Acceleration

2. Force

3. Half

4. Balanced

5. Different

6. Third

7. (a) Newton's First law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

Example: When the playing card is flicked with the finger the coin placed over it falls in the tumbler.

(b) Second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Example: In a high jump athletic event, the athletes are made to fall either on a cushioned bed or on a sand bed. This is to increase the time of the athlete’s fall to stop after making the jump. This decreases the rate of change of momentum and hence the force.

(c) Third law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies.

The third law of motion can also be illustrated when a sailor jumps out of a rowing boat. As the sailor jumps forward, the force on the boat moves it backwards

8. Three types of inertia are:

(a) Inertia of rest: The ability of a body to resist any change in its state of rest. Example: It is a common experience to have a jerk when a vehicle starts moving from its position of rest.

(b) Inertia of motion: The ability of a body to resist any change in its state of motion. Example: The passenger in a running bus tends to lean forward, when the bus stops suddenly

(c) Inertia of rotation: The ability of a body to resist any change in its state of rotation is inertia of rotation. Example: Think of a metal ring and a solid disk experiment. The metal ring has its mass at the perimeter and therefore has more rotational inertia than the solid disk.

9. Principle states that the linear momentum of a system has constant magnitu deand direction ift he system is subject edton oexternal force.

Derivation:

F = dp/dt

F=(pf-pi)/dt

When force in zero

= Pf- pi= 0

= Pf= pi

Hence momentum remains same.

10. (i) Momentum is a vector

Because we obtain momentum by multiplying mass m (a “number”) and velocity v, the end result is also a vector. i.e. it has a magnitude and direction.

(ii) Momentum of an object is parallel to the object’s velocity. Multiplication of v by m changes the magnitude and units, but as m is a number it has no information about direction. Because masses are always positive, p and v for the same object always point the same way.

(iii) The SI units of momentum are kg m/s

11. M = 1kg, u = 15m/s, v = 25ms, t = 5sec

v = u+d

25 = 15+a x 5

10 = 5a

a = 2ms2

F = ma

F = 1 x 2

= 2 newbons

12. Mshell = 6kg, mgrm = 600kg, vgrm = 3m/s, vshell= ?

using conservator of momentum

Pi = Pf

0 = mgvi + mshell v2

0 = 600 x 3 + 6 x vshell

vshell =-1800/6

= -300m/s

13. m = 20kg, u = 0, F = 10N, t=10 second,

F = ma

10 = 20 x a

a = 10/20 = 0.5m/s2

v = u + at

v = 0 + 0.5 x 10

v = 5ms

5 = at + 1/2at2

5 = 0 x 10 + 1/2 x 0.5 x (10)2

5 = 0 + 1/2 x 1/2 x 100

5 = 25 meters

14. m1 = 1kg, u1 = 4m/s, v1 = 8/3 m/s

m2 = 0.5 kg, u2 = 2m/s, v2 = ?

conservatin of momentum

pi = pf

m1u1 + m2u2 = m1v1 + m2v2

(1 x 4) + (0.5 x 2) = (1 x 8/3) + (0.5 x v2)

4+1 = 8/3 + v2/2

5 - 8/3 = v2/2

15-8/3 = v2/2

7 x 2/ 3 = v2

14/3 = v2

v2 = 14/3 m/s

15. Mb = 10g, ub = 1.5m/s, vb = vw = ?

Mw = 90, uW = 0

Using conservation of momentum

mbub + m2uw) = mbvb + mwvw = v(mb + 10nW)

(10 x 1.5) + (90 x 0) = (10 + 90)v

15 + 0 = 100v

v = 15/100 = 0.15m/s