There are many kinds of forces in nature – the effects of any forces are described accurately by three general laws of motion formulated by Sir Isaac Newton.
According to Newton’s First Law of Motion, “Every body continues to be in a state of rest or of uniform motion along a straight line unless it is compelled to change that state by an external agency acting on it.”
Newton’s Second Law of Motion states that “The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.”
Newton’s Third Law of Motion states that “ For every action, there is an equal and opposite reaction.”
Download JEE Advanced Previous Year Questions with Solutions on Newton’s law of motion PDF
Question 1) A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes up to 2 m height further, find the magnitude of the force. (take, g =10 ms-2.)
1) 4 N
2) 16 N
3) 20 N
4) 22 N
Answer: 4) 22 N
Solution:
Consider the equation v2 – u2 =2as for the upward journey
Final velocity of the ball, v = 0
⇒ -u2 = 2(-10) x 2
u2 = 40
Again applying v2 – u2 =2as for the ball in the hands of the thrower
Initial velocity, u =0
⇒ 40 – 0= 2(a)0.2
a = 100 m/s2
Applied force, F = mg + ma
= (0.2)(10) + (0.2)(100)
= 2 + 20
= 22 N
Question 2) A force of 1200 N acts on a 0.5 kg steel ball as a result of a collision lasting 25 ms. If the force is in a direction opposite to the initial velocity of 14 ms-1, then the final speed of the steel ball would be
1) 24 ms-1
2) 35 ms-1
3) 12 ms-1
4) 46 ms-1
Answer: 4) 46 ms-1
Solution:
For an impulsive force of 1200 N
F = Δp/Δt
F (Δt) = Δp
– 1200 (t2 – t1) = P2 – P1
– 1200 x 25 x 10-3 = -mv’ – mv
1200 x 25 x 10-3 = (1/2) v’ + (1/2)v
2 x 1200 x 12 x 10-3 = v’ + v
2 x 1200 x 12 x 10-3 = v’ + 14
60 = v’ + 14
v’ = 60 – 14
= 46 m/s
Question 3) A stationary bomb explodes into three pieces. One piece of 2 kg mass moves with a velocity of 8ms-1 at right angles to the other piece of mass 1 kg moving with a velocity of 12 ms-1. If the mass of the third piece is 0.5 kg, then its velocity is
1) 10 ms-1
2) 20 ms-1
3) 30 ms-1
4) 40 ms-1
5) 50 ms-1
Answer: 4) 40 ms-1
Solution:
Initial momentum is zero since the ball is stationary
Momentum of 2kg is P1 = 2 x 8 = 16 kgm/s
Momentum of 1 kg is P2 = 1 x 12 = 12 kgm/s
Both are at right angles so the resultant will be P = √P12 + √P22
= √256 + √144 = 20 kgm/s
the third piece will have a momentum in the opposite direction to make the net momentum zero.
Therefore V = P/m = 20/0.5 = 40 m/s
Question 4) A cricket ball of mass 0.25 kg with a speed of 10 m/s collides with a bat and returns with the same speed within 0.01s. The force acting on the bat is
1) 25 N
2) 50 N
3) 250 N
4) 500 N
Answer: 4) 500 N
Solution:
Δp = 2mv = 2 x 0.25 x 10 = 5 kgm/s
F = Δp/Δt = 5/0.01 = 500 N
Question 5) A hockey player receives a corner shot at a speed of 15 ms-1 at an angle of 30° with the y – axis and then shoots the ball of mass 100 g along the negative x-axis with a speed of 30 ms-1. If it remains in contact with the hockey stick for 0.01s, the force imparted to the ball in the x-direction is
1) 281.25 N
2) 187.5 N
3) 562.5 N
4) 375 N
Answer: 4) 375 N
Solution:
Mass of the ball, m = 100 g = 0.1 Kg
Initial velocity of the ball, u = 30 m/s
Final velocity of the ball, v = 15 m/s
Change in momentum of ball along x- axis ΔP = mv – m(-usin 300)
ΔP = mv + musin 300
= (0.1 x 30) + (0.1 x 15 x 0.5)
= 3 + 0.75
= 3.75
Force exerted by the ball = 3.75/0.01 = 375 N
Question 6) A rocket of mass 100 kg burns 0.1 kg of fuel per second. If the velocity of exhaust gas is 1 kms-1 , then it lifts with an acceleration of
1) 1000 ms-2
2) 100 ms-2
3) 10 ms-2
4) 1 ms-2
Answer: 4) 1 ms-2
Solution:
Mass (m) = 100 Kg
Velocity= 1 km/sec = 1000 m/s
dm/dt = 0.1
Thrust force, F = V. (dm/dt)
= 1000 x 0.1= 100
Now,
ma = F – mg
100 x a = 100 – m x 0
a = 100/100
= 1 m/s2
Question 7) A bullet of mass 10 g moving with 300 m/s hits a block of ice of mass 5 kg and drops dead. The velocity of ice is
1) 50 cm/s
2) 60 cm/s
3) 40 cm/s
4) 200 cm/s
Answer: 2) 60 cm/s
Solution:
Applying law of conservation of linear momentum
m1u1+m2u2=m1v1+m2v2
10-2 x 300 + 5 x 0 = 10-2 x 0 + 5v2
5v2= 3
v2= (3/5) m/s = 60 cm/s
Question 8) 10000 small balls, each weighing lg, strike one square centimetre of area per second with a velocity 100 ms-1 in a normal direction and rebound with the same velocity. The value of pressure on the surface will be
1) 2 × 103 Nm-2
2) 2 × 105 Nm-2
3) 107 Nm-2
4) 2 × 107 Nm-2
Answer: 4) 2 × 107 Nm-2
Solution:
Pressure = Force/Area
= n(mv- (-mv))/A
= 2mnv/A
= (2 x 10-3 x 104x 102)/10-4
= 2 x 107 N/m2
Question 9) A block of mass 2 kg is at rest on a floor. The coefficient of static friction between block and the floor is 0.54. A horizontal force of 2.8 N is applied to the block. What should be the frictional force between the block and the floor? (take, g =10 m/s2)
1) 8.8 N
2) 5.8 N
3) 2.8 N
4) 10.8 N
Answer: 3) 2.8 N
Solution:
mass of block = 2kg
coefficient of static friction = 0.4
Force applied to the block, Fnet = 2.8 N
The force of friction, f = μN
μ is the coefficient of static friction
N is the normal force
N=mg
f = μmg
= 0.4 x 2 x 10
= 8 N
Question 10) Starting from rest, the time taken by a body sliding down on a rough inclined plane at 450 with the horizontal is twice the time taken to travel on a smooth plane of the same inclination and same distance. Then, the coefficient of kinetic friction is
1) 0.25
2) 0.33
3) 0.50
4) 0.75
Answer: 4) 0.75
Solution:
Coefficient of kinetic friction, μ = tanθ[1 – 1/n2]
here θ = 450
and n=2
μ = tan 450[1 – 1/22]
= 1 – 1/4
= 3/4
= 0.75
Recommended Video
Comments