## A bomb of 1kg is thrown vertically up with speed 100 m/s. After 5 seconds, it explodes into two parts…

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.

## 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…

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

## A particle of mass 4 m explodes into three pieces of masses m,m and 2m. The equal masses move along X-axis and Y-axis…

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

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

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

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

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

## A bomb moving with velocity $(40 \hat{i} + 50 \hat{j}-25 \hat{k}) m/s$ explodes into two pieces of mass ratio…

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 $\large \vec{v_2} = 45 \hat{j} – 35 \hat{k}$

## The maximum tension a rope can withstand is 60 kg-wt. The ratio of maximum acceleration with which two boys of masses…

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.

## 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…

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

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

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

## A mass of 1 kg attached to one end of a string is first lifted up with an acceleration 4.9m/s^2 and then lowered with same acceleration…

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

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