Question

A girl of mass 40 kg jumps with a velocity of 5 m/sec stationery cart with a frictionless wheel the mass of cart 3 kg what is the velocity of cart start moving.​

Answer 1

Given:-

• Mass of girl ( m¹ ) = 40kg

• Initial Velocity of girl ( u¹ ) = 5m/s

• Mass of Cart ( m² ) = 3kg

• Initial Velocity of Cart ( u² ) = 0m/s

To Find:-

• The Velocity with which cart starts

Formulae used:-

• Conservation of Momentum

Now,

As the girl jumps onto the cart,they would have same Velocity while moving.

So, v¹ = v²

Therefore,

→ m¹u¹ + m²u² = m¹v¹ + m²v²

→ 40 × 5 + 3 × 0 = (m¹ + m²)v

→ 200 = ( 40 + 3)v

→ 200 = 43v

→ v = 200/43

→ v = 4.65m/s

Hence, The cart will start with the Velocity of 4.65m/s

Answer 2

$\large{\bf{\red{\underline{\underline{Given }}}}}$

• Mass of a girl (m¹)=40kg
• Initial velocity of a girl (u¹) =5m/s
• Mass of cart ( m²) = 3kg
• Initial velocity of cart ( u² ) = Om/s

$\large{\bf{\purple{\underline{\underline{To \:find : }}}}}$

• The velocity which cart starts .

$\large{\sf{\red{\underline{\underline{Formula \:Used}}}}}$

• Conversation of Mòmèñtum

$\boxed{ \huge{ \fcolorbox{yellow}{white}{ \pink{~Now~}}} }$

• As the girl jumps onto to the card, they would have same velocity of moving.

So , v¹ = v²

$\large{\bf{\blue{\underline{\underline{Therefore,}}}}}$

$= > {m}^{1} {u}^{2} = {m}^{1} {v}^{1} + {m}^{2} {v}^{2}$

$= > 40 \times 5 + 3 \times 0 = ( {m}^{1} + {m}^{2} )v$

$= > 200 = (40 + 3)v$

$= > v = \frac{200}{3}$

$= > v = 4.65m/s$

The cart will be start at the velocity of 4.65m/s.