Question

A girl of mass 40 kg jumps with a horizontal velocity of 5 m s-1 onto a stationary cart with friction less wheels.The mass of the cart is 3kg.
What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction??​

Answer 1

Answer:

$\huge\bold\star\fbox\red{Question ✿ ࿐}$

A girl of mass 40 kg jumps with a horizontal velocity of 5 m s-1 onto a stationary cart with friction less wheels.The mass of the cart is 3kg.

What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction??

$\huge\bold\star\fbox\red{Answer ✿ ࿐}$

Let v be the velocity of the girl on the cart as the cart moving.

The total momenta of the girl and cart before the interaction

$= 40kg \: \times 5m \: s {}^{ - 1} + 3kg \times 0 \: \: m \: s \frac{ - 1}{ } \\ 200 \: \: kg \: \: m \: \: s \frac{ - 1}{ }$

Total Momenta After The Interaction

$(40 + 3)kg \times v \: \: kg \: \: m \: \: s \frac{ - 1}{}.$

According To The Law Of Conservation Of Momentum,

the total momentum is conserved during the Interaction.

That Is,

$43 \: \: v \: \: = 200 \\ ⟹ \: v \: = \begin{gathered} \rm { \cancel{\dfrac{200}{43}} = }\\ \end{gathered}4.65 \: m \: \: s \frac{ - 1}{}.$

The girl on cart would move with a velocity of

$4.65 \: \: m \: \: s \frac{ - 1}{}.$

in the direction in the which the girl jumped