1)
A body of mass 3-kg is subjected to mutually perpendicular
forces 4 N and 6 N. Find magnitude and direction of acceleration
of the body.
Solution :
C
B


Let P = 4N, Q = 6N act on the body
At right angles to each other
Q = 6N
θ
0
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P = 4N
C
B
Solution Continued
Q = 6N
θ
0
P = 4N
Given m=3kg
Find the acceleration acting on
the body due to the two forces.
Now find the magnitude of the
acceleration of the body along OC.
from F = ma, a = F = 7.211= 2.4 ms -2
m
3
The magnitude of the acceleration of the

body along OC is R.


Let θ be the angle between P and R
tanθ = Q = 6 = 3 = 1.5 or θ = 56o18'.
P 4 2
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2) A body of mass 0.2 kg which is at rest is acted upon by a
force 8N for 4 s. Find
(i) the change in momentum of the body and
(ii) the velocity of the body.
Initial velocity, u = 0;
Solution :
t=4s
∴ Initial momentum, mu=0;
Let the final velocity = v ;
To find the change in momentum
take the difference of final and
initial momenta.
∴Final momentum = mv
∴Change in momentum = mv – 0=mv
∴But change in momentum is nothing
but the impulse (F t)on the body.
∴ change in momentum = Ft
=8x4
= 32 N-s
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v=?
m= 0.2 kg
8N
t = 4s
Solution Continued
m= 0.2 kg
v=?
8N
t = 4s
Now find the velocity of the
body for this we can use the
equation Ft = mv
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∴v
= Ft = 32 = 160 m/s
m 0.2
3) A body of mass 16 kg is moving with a velocity 1.5 ms-1
calculate the force required to stop it in 8 s. Find the distance
travelled by the body before coming to rest.
Solution :
Given, m=16kg, u=1.5 ms-1 t=8s,
v=0, F=? s=?
F=?
M = 16 kg
We know F = m(v - u) = 16(0 - 1.5) =-3N
t
8
Negative sign indicates that the
force should be applied in a
direction opposite to the
direction of
motion of the body. a = F = -3 = -0.1875
m 16
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A
B
4 = 1.5 m/s
Within
8s
S=?
Solution Continued
F=?
M = 16 kg
A
B
4 = 1.5 m/s
Within
8s
S=?
Next we are asked to find the
distance travelled by the body
before coming to rest, i.e., s.
To find the value of s we require
the value of because a appears
in the two equations of motion
that contain s.
Now u,v and a are known. s can be
calculated by v2=u2+2as
©Bangalore Institute Of Coaching
From F=ma, we get
a = F = -3 = - 0.1875 ms -2
m 16
2
2 -u2
0.(15)
v
s=
=
= 6m
2a
12* - 0.1875
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4) A foot ball weighing 0.25 kg is kicked by a player.
The ball moves with a speed 30 ms-1. It is brought to rest by
another player in 0.04 s. Calculate the impules of the ball on
the second player who stopped it and the average force used by him.
Solution :
Given m=0.25 kg. u=30m/s,
v=0, t=0.04 s, F=? F=?
Impulse,
I = Ft=m(v-u)=0.25(0.30)=-7.5 N.
negatively sigh indicated that the
velocity is decreased.
Find the average Force.
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Average force, F = l = 7.5 = 187.5 N.
t 0.04
5) A shot put of mass 5kg falls freely under gravity.
Find (i) the force acting on it
(ii) the momentum acquired by it after 5 s and
(iii) the velocity after 4s. Given, g=9.8 ms-2
Solution :
Given, m=5kg, u=0, g=9.8s-2,
F=? P=? v=?
Find the force acting on it.
©Bangalore Institute Of Coaching
Force acting on shotput
F=mg=5*9.8=49 N.
Solution Continued
Find the momentum acquired
by it after 5 s.
Find the velocity after 4 s.
Given, g=9.8 ms-2
©Bangalore Institute Of Coaching
momentum acquired after 5 s.
To find the momentum after 5 s.,
we have to first find the
velocity after 5 s.
v=u+gt=0+9.8 *5=49 ms-1
Velocity after 4 s.
change in momentum =mvmu=0=mv=Ft
∴
v = Ft = 49*4 = 39.2ms -1
m
5
6) Two solid objects of mass 8 kg and 18 kg are moving with
velocities 18ms-1 and 4ms-1 respectively in the same direction.
They collide at some point After collision, the ligher of the
two moves with velocity 6ms-1 in the same direction .
What is the velocity of the heavier object?
Solution :
m1=8 kg, m2=18 kg, u1=14m/s,
u2=4m/s, v1=6 m/s, v2=?
m1u1+m2u2=m1v1+m2v2
Use the law of conservation of
momentum.
8*14+18*4=8 * 6+18* v
112+72=48+18 *v
Or,
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2
2
v2 = 184 - 48 = 136 = 7.55 m/s
18
18