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:

Step-by-step explanation:

Here, m1=60kg,u1=3m/s,m2=140kg,u2=1.5m/s

if v is their common velocity, then applying the principle of conservation of linear momentum, we get

m1u1+m2u2=(m1+m2)v or 60×3+140×1.5=(60+140)v

or 180+210=200v or v=390/200=1.95m/s

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Answer 2

Provided that:

• Mass of boy = 60 kg

• Initial velocity of boy = 3 m/s

• Mass of trolley = 140 kg

• Initial velocity of trolley = 1.5 m/s

To calculate:

• With what velocity will the two moves together, in short we have to find out their combined velocity!

Solution:

• The both moved with a velocity of 1.95 metre per second.

Using concept:

• Law of conservation of momentum

Using formula:

${\small{\underline{\boxed{\sf{\rightarrow \: m_A u_A + m_B u_B \: = m_A v_A + m_B v_B}}}}}$

(Where, ${\sf{m_A}}$ denotes mass of object one, ${\sf{u_A}}$ denotes initial velocity of object one, ${\sf{m_B}}$ denotes mass of object two, ${\sf{u_B}}$ denotes initial velocity of object two, ${\sf{v_A}}$ denotes final velocity of object one, ${\sf{v_B}}$ denotes final velocity of object two.)

Required solution:

$:\implies \sf m_A u_A + m_B u_B \: = m_A v_A + m_B v_B \\ \\ :\implies \sf 60(3) + 140(1.5) = (60+140)v \\ \\ :\implies \sf 180 + 210 = 200v \\ \\ :\implies \sf 390 = 200v \\ \\ :\implies \sf \dfrac{390}{200} \: = v \\ \\ :\implies \sf \dfrac{39}{20} \: = v \\ \\ :\implies \sf 1.95 \: = v \\ \\ :\implies \sf v \: = 1.95 \: ms^{-1} \\ \\ :\implies \sf Combined \: velocity \: = 1.95 \: ms^{-1}$