Question

A body of mass 100g moves at a speed 2m/s,And another body of mass 200g moves at a speed 1m/s,in a straight line and in the same direction.After collision both bodies moves at the same speed.If so calculate the speed of the bodies?​

Answer 1

$\\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

• m(1) = 100g = 0.1kg

• m(2) = 200g = 0.2kg

• u(1) = 2m/s

• u(2) = 1m/sec

• Final velocity (v) is same for both the body.

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• The final velocities of the bodies.

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

Total momentum of the body is conserved as no external force is applied to the system.

So,

$\\ \small \boxed{ \bf \: m(1)u(1) + m(2)u(2) = m(1)(v1) + m(2)v(2)}$

Here ,

v(1) = v(2) = v

Putting values , we get...

➠ 0.1(2) + 0.2(1) = 0.1(v) + 0.2(v)

➠ 0.2 + 0.2 = 0.1v + 0.2v

➠ 0.4 = 0.3v

➠ v = 0.4/0.3

➠ v = (4/3) m/sec or 1.33 m/sec.

Hence , final speed of the bodies is 4/3 m/sec or 1.33m/sec.

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NOTE :-

• In the question, mass is given in grams(g). So , we have to convert it to kilo grams(kg).

1000g = 1kg

Answer 2

Let us assume that there is no external unbalanced force acting in the system.

Formula of law of conservation of Momentum:-

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Before collision:-

m₁u₁ + m₂u₂

= (100 g)(2 m/s) + (200 g)(1 m/s)

= 200 g m/s + 200 g m/s

= 400 g m/s ...(A)

After collision:-

Let, final velocity = v

v(m₂ + m₁) {Given}

= v(100 g + 200 g)

= 300 g × v ...(B)

∵ A = B,

∴ 300 g × v = 400 g m/s

⇒300v = 400 m/s

⇒v = 4/3 m/s

⇒v = 1.33 m/s {Speed of both the bodies after collision}