Question

A boy of mass 60 kg running at 3 m/s jumps on to a trolley of mass 140 kg moving with
velocity of 1.5 m/s in the same direction. With what velocity will the two moves
together?

Answer 1

Answer :-

The boy and the trolley will move together with a velocity of 1.95 m/s .

Explanation :-

We have :-

→ Mass of the boy (m₁) = 60 kg

→ Velocity of the boy (u₁) = 3 m/s

→ Mass of the trolley (m₂) = 140 kg

→ Velocity of the trolley (u₂) = 1.5 m/s

To find :-

→ Their combined velocity (v) .

________________________________

As the boy and the trolley were moving in the direction .

So their combined velocity according to "Law of Conservation of Momentum" will be :-

m₁u₁ + m₂u₂ = (m₁ + m₂)v

⇒ 60 × 3 + 140 × 1.5 = (60 + 140)v

⇒ 180 + 210 = 200v

⇒ 390 = 200v

⇒ v = 390/200

⇒ v = 3.9/2

⇒ v = 1.95 m/s

Answer 2

Answer:

Given :-

• A boy of mass 60 kg running at 3 m/s jumps on to a trolley of mass 140 kg moving with velocity of 1.5 m/s in the same direction.

To Find :-

• What is the velocity will the two moves together.

Formula Used :-

$\clubsuit$ Momentum Formula :

$\mapsto \sf\boxed{\bold{\pink{m_1u_1 + m_2u_2 =\: (m_1 + m_2)v}}}$

Solution :-

Given :

• Mass of boy (m₁) = 60 kg
• Velocity of boy (u₁) = 3 m/s
• Mass of trolley (m₂) = 140 kg
• Velocity of trolley (u₂) = 1.5 m/s

According to the question by using the formula we get,

$\longrightarrow \sf (60)(3) + (140)(1.5) =\: (60 + 140)v$

$\longrightarrow \sf 60 \times 3 + 140 \times 1.5 =\: (200)v$

$\longrightarrow \sf 180 + 140 \times \dfrac{15}{10} =\: 200v$

$\longrightarrow \sf 180 + \dfrac{210\cancel{0}}{1\cancel{0}} =\: 200v$

$\longrightarrow \sf 180 + 210 =\: 200v$

$\longrightarrow \sf 390 =\: 200v$

$\longrightarrow \sf \dfrac{39\cancel{0}}{20\cancel{0}} =\: v$

$\longrightarrow \sf \dfrac{39}{20} =\: v$

$\longrightarrow \sf 1.95 =\: v$

$\longrightarrow \sf\bold{\red{v =\: 1.95\: m/s}}$

${\small{\bold{\underline{\therefore\: The\: velocity\: will\: the\: two\: moves\: together\: is\: 1.95\: m/s\: .}}}}$