Laws of Motion

Q: A bomb of 1 kg is thrown vertically up with speed 100 m/s. After 5 seconds, it explodes into two parts. One part of mass 400 gm goes down with speed 25 m/s. What will happen to the other part just after explosion

Sol: After 5 sec,  velocity of the bomb,

$\large \vec{v} = \vec{u} + \vec{gt}$

$\large \vec{v} = u \hat{j} – g t \hat{j} $

$\large \vec{v} = 100 \hat{j} – 10 \times 5 \hat{j} $

$\large \vec{v} = 50 \hat{j} m/s$

m = 1kg , m₁ = 0.4 kg, m₂ = 0.6 kg, v₁ = 25 m/s

According to law of conservation of momentum

$\large \vec{m v} = \vec{m_1 v_1} + \vec{m_2 v_2} $

$\large 1 \times 50 \hat{j} = -0.4 \times 25 \hat{j} + 0.6 \vec{v_2}$

$\large \vec{v_2} = 100 \hat{j}$ ; vertically upwards.

Q: A rifle of 20 kg mass can fire 4 bullets/s. The mass of each bullet is 35 × 10^-3 kg and its final velocity is 400 m/s. Then, what force must be applied on the rifle so that it does not move backwards while firing the bullets ?

Sol: Law of conservation of momentum $\large \vec{M V} + \vec{4m v} = 0 $

$\large \vec{V} = -\frac{\vec{4m v}}{M} $

$\large = -\frac{4 \times 35 \times 10^{-3}\times 400}{20} $

= -2.8 m/s

Force applied on the rifle

$\large F = \frac{MV}{t} = -\frac{20 \times 2.8}{1} $

= -56 N

Q: A particle of mass 4 m explodes into three pieces of masses m , m and 2 m. The equal masses move along X-axis and Y-axis with velocities 4 m/s and 6 m/s respectively. The magnitude of the velocity of the heavier mass is

Sol: M = 4m , u = 0 , m₁ = m, m₂ = m, m₃ = 2m

v₁ = 4m/s, v₂ = 6 m/s, v₃ = ?

According to law of conservation of momentum,

$\large \vec{p_1} + \vec{p_2} + \vec{p_3} = 0 $

$\large \vec{p_3} = -(\vec{p_1} + \vec{p_2}) $

$\large p_3 = \sqrt{p_1^2 + p_2^2 + 2 p_1 p_2 cos\theta}$

Since , P₁ and P₂ are perpendicular to each other

$\large p_3 = \sqrt{p_1^2 + p_2^2 }$

$\large m_3 v_3 = \sqrt{(m_1 v_1)^2 + (m_2 v_2)^2 }$

$\large 2m v_3 = \sqrt{(m \times 4)^2 + (m \times 6)^2 }$

$ v_3 = \sqrt{13} m/s $

Q: A bomb moving with velocity $ (40 \hat{i} + 50 \hat{j}-25 \hat{k}) m/s $ explodes into two pieces of mass ratio 1:4 after explosion the smaller piece moves away with velocity $ (200 \hat{i} + 70 \hat{j} + 15 \hat{k}) m/s $ . The velocity of larger piece after explosion is

Sol: From law of conservation of linear momentum

M u = m₁ v₁ + m₂ v₂ ;

M = 5m , m₁ = m , m₂ = 4m

$\large \vec{u} = (40 \hat{i} + 50 \hat{j}-25 \hat{k}) m/s $

$\large \vec{v_1} = (200 \hat{i} + 70 \hat{j} + 15 \hat{k}) m/s $

Let = v₂ is the velocity of the larger piece

$\large 5m (40 \hat{i} + 50 \hat{j}-25 \hat{k}) = m(200 \hat{i} + 70 \hat{j} + 15 \hat{k}) + 4m (v_2) $

on simplification, we get $\large \vec{v_2} = 45 \hat{j} – 35 \hat{k} $

Q: The maximum tension a rope can withstand is 60 kg-wt. The ratio of maximum acceleration with which two boys of masses 20 kg and 30 kg can climb up the rope at the same time is

Sol: m₁ = 20 kg, m₂ = 30 kg, T = 60 kgwt = 600 N

For ‘m₁ ’ ; T – m₁ g = m₁ a₁

600 – 20 × 10 = 20 × a₁

⇒ a₁ = 20 ms^-2

For ‘m₂ ’ ; T – m₂ g = m₂ a₂

600 – 30 × 10 = 30 × a₂

⇒ a₂ = 10 ms^-2

a₁ : a₂ = 20:10 = 2:1.

Q: The apparent weight of a man in a lift is W₁ when lift moves upwards with some acceleration and is W₂ when it is accelerating down with same acceleration. Find the true weight of the man and acceleration of lift.

Sol: (a) W₁ = m(g + a),

W₂ = m(g – a)

W₁ + W₂ = 2 mg

⇒ W₁ + W₂ = 2 W (∵ W = mg)

$\large W = \frac{W_1 + W_2}{2}$

(b) $\large \frac{W_1}{W_2} = \frac{g+a}{g-a} $

$\large \frac{g}{a} = \frac{W_1 + W_2}{W_1 – W_2}$

$\large a = g( \frac{W_1 – W_2}{W_1 + W_2} ) $

Q: A mass of 1 kg attached to one end of a string is first lifted up with an acceleration 4.9 m/s² and then lowered with same acceleration. What is the ratio of tension in string in two cases.

Sol: When mass is lifted up with acceleration 4.9 m/s²

T₁ = m(g + a) = 1 (9.8 + 4.9) = 14.7 N

When mass is lowered with same acceleration

T₂ = m(g – a) = 1 (9.8 – 4.9) = 4.9 N

$\large \frac{T_1}{T_2} = \frac{14.7}{4.9} = 3:1 $