Chapter 4 Laws of Motion Solutions

Maharashtra State Board 11th Physics Solutions Chapter 4 Laws of Motion

1. Choose the correct answer.

2. Answer the following questions.

Thus, out of the four fundamental forces, electromagnetic force governs our daily life almost entirely.

In Chapter 5, you will study about force of gravitation in detail.

(B) Conservative and non-conservative forces,

(C) Contact and non-contact forces,

(D) Inertial and non-inertial frames of reference.

Conditions/limitations for application of work formula:

Integral method to find work done by a variable force:

Thus, work and energy are the two sides of the same

i. For two colliding bodies, the negative of ratio of relative velocity of separation to relative velocity of approach is called the coefficient of restitution.

ii. Consider an head-on collision of two bodies of masses m₁and m₂with respective initial velocities u₁and u₂. As the collision is head on, the colliding masses are along the same line before and after the collision. Relative velocity of approach is given as,
u_a= u₂– u₁
Let v₁and v₂be their respective velocities after the collision. The relative velocity of recede (or separation) is then v_s= v₂– v₁

iii. For a head on elastic collision, According to the principle of conservation of linear momentum,
Total initial momentum = Total final momentum

iv. For a perfectly inelastic collision, the colliding bodies move jointly after the collision, i.e., v₁= v₂
∴ v₁– v₂= 0
Substituting this in equation (1), e = 0.

ii. A very heavy object collides with a lighter object, initially at rest.
Let m₁be the mass of the heavier body and m₂be the mass of the lighter body i.e., m₁>> m₂; lighter particle is at rest i.e., u₂= 0 then,

i.e., the heavier colliding body is left unaffected and the lighter body which is struck, travels with double the speed of the massive striking body.

iii. A very light object collides on a comparatively much massive object, initially at rest.
If m₁is mass of a light body and m₂is mass of heavy body i.e., m₁<< m₂and u₂= 0. Thus, m₁can be neglected.
Hence v₁≅ -u₁, and v₂≅ 0.
i.e., the tiny (lighter) object rebounds with same speed while the massive object is unaffected.

Loss in the kinetic energy during a perfectly inelastic head on collision:

Need to define impulse:

Impulse for a variable force:

In chapter 3, you have studied the concept of using area under the curve.

3. Solve the following problems.

Numerical Analysis

i. Consider FBD for 20 kg-wt load. Initially, the block kept on the table is moving towards left, because of the movement of block of mass 20 kg in downward direction.
Thus, for block of mass 20 kg,
ma = mg – T₁…. (1)
Consider the forces acting on the block of mass 35 kg in horizontal direction only as shown in figure (b). Thus, the force equation for this block is, m₁a = T₁– T₂– f ….(2)
To prevent the block from sliding across the table,
m₁a = ma = 0
∴ T₁= mg = 200 N ….[From (1)]
T₁= T₂+ f ….[From (2)]
∴ T₂+ f = 200
∴ T₂= 200 – 35 = 165 N
Thus, the total force acting on the block from right hand side should be 165 N.
∴ Total mass = 16.5 kg
∴ Minimum weight to be added = 16.5 – 2 = 14.5 kg
≈ 15 weights of 1 kg each

ii. Now, considering motion of the block towards right, the force equations for the masses in the pan and the block of mass 35 kg can be determined from FBD shown

From figure (c)
m₁a = T₂– T₁– f ….(iii)
From figure (d),
m₂a = m₂g – T₂….(iv)
To prevent the block of mass 35 kg from sliding across the table, m₁a = m₂a = 0
From equations (iii) and (iv),
T₂= T₁+ f
T₂= m₂g
∴ m₂g = 200 + 35 = 235 N
∴ The maximum mass required to stop the sliding = 23.5 – 2 = 21.5kg ≈ 21 weights of 1 kg
Answer:
The minimum 15 weights and maximum 21 weights of 1 kg each are required to stop the block from sliding.

i. A constant force F is applied to a body of mass (m) initially at rest (u = 0).

ii. We have,
v = u + at
∴ v = 0 + at
∴ v = at …. (1)

iii. Now, power is the rate of doing work,
∴ P =
∴ P = F. [∵ dW = F. ds]

iv. But = v, the instantaneous velocity of the particle.
∴ P = F.V … (2)

v. According to Newton’s second law,
F = ma … (3)

vi. Substituting equations (1) and (3) in equation (2)
P = (ma) (at)
∴ P = ma²t
∴ P = × t
∴ P =

vii. As F and m are constant, therefore, P ∝ t.

Alternate solution:
Work done, w = Area of trapezium ADCB
∴ W = (AD + CB) × DC
∴ W = 1 (5N + 3N) × (5m – 3m)
= × 8 × 2 = 8J

ii. From equation (1),
∴ v = – eu
∴ After first bounce,
v₁= – eu
after second bounce,
v₂= -ev₁= -e(-eu)= e²u
and after third bounce,
v₃= – ev₂= – e(e²u) = – e³u
But u =

iii. Impulse given by the ball during third bounce, is,

iv. Average force exerted in 250 ms,

v. Average pressure for area
0.5 cm²= 0.5 × 10^-4m²
P = = 9.216 × 10⁴N/m²

i. m₁u₁+m₂u₂= m₁v₁+ m₂v₂
ii. e =
Calculation: From formula (i),
[(2m) × 6] + (m × 12) = (2m × 9) + mv₂
∴ v₂= 6cm/s

From formula (ii),
e = = = 0.5

Answer: The coefficient of restitution is 0.5

The ladder exerts horizontal force on the wall at A and is the force exerted on the ground at C.
As || = … (iii)

Let monkey climb upto distance x along BC (Horizontal) i.e., CM’ = x....(iv)
Then, the net normal reaction at point C will be, N = 100 + 200 = 300N
From equation (iii),
H = = 150N
By condition of equilibrium, taking moments about C,
(-H × AB) + (W₁× CD’) + (W₂× CM’) + (F × 0)’0
∴ (-150 × 1.6) + (100 × 0.6) + (200 × x) = 0
∴ 60 + 200x = 240
∴ 200x = 180
∴ x = 0.9

From figure, it can be shown that,
∆ABC ~ ∆MM’C
∴ = ∴ =
∴ CM = 1.5 m

Answer:

11th Physics Digest Chapter4 Laws of MotionIntext Questions and Answers

Can you recall? (Textbook Page No. 47)

Can you tell? (Textbook Page No. 48)

Can you tell? (Textbook Page No. 48)