Question

a girl of mass 40 kg jumps with a horizontal velocity of 5m/s-1into a stationary cart with frictionless wheels the mass of the cart is 3kg . what is her velocity as the cart is stark moving ? assume that there is no external unbalanced force working in the horizontal direction

Answer 1

Explanation:

Given:-

• Mass of the girl $\rm (m_{1})$ = 40kg

• Mass of cart $\rm (m_{2})$ = 3kg.

• Velocity of the girl $\rm (u_{1})$ = 5m/s

• Velocity of the cart $\rm (u_{2})$ = 0m/s

To Find:-

• The velocity of the girl after the cart starts moving.

Solution:-

Since, the girl was on the cart.

Total mass = Mass of girl + Mass of cart

= 40 + 3

= 43kg

Let the velocity of the girl after the cart starts moving be y

Momentum of the girl and cart before interaction:-

= Momentum of girl + Momentum of cart

$\sf = (m_{1} \times u_{1}) + (m_{2} \times u_{2})$

$\sf = (40\times 5 ) + (3 \times 0)$

$\sf = 200 + 0$

$\sf = 200$

Momentum of girl and cart before interaction = 200kg m/s

Momentum of the girl and cart after interaction:-

= Total mass × Velocity after the cart starts moving

= 43 × v

= 43v

Momentum of the girl and cart after interaction = 43v

According the the law of conservation of momentum:-

Momentum before the interaction = Momentum before the interaction

$\sf\dashrightarrow 200 = 43v$

$\sf\dashrightarrow v = \dfrac{200}{43}$

$\sf\dashrightarrow v = 4.65 m/s$

Velocity of the girl after the cart starts moving = 4.65 m/s

Answer 2

Answer:

Mass of the girl = 40 kg

Mass of the cart = 3 kg

Velocity of the girl = 5 m/s

Velocity of the cart = 0 m/s

To Find :-

Velocity as the cart starts moving.

Solution :-

Firstly let's find the Mounmentum of girl

[P = mv]

Here,

P = Mounmentum

M = Mass

V = Velocity

P = 40 × 5

Hence, Mounmentum of girl is 200 kg m/s .

Now,

Let's add the Mounmentum of girl and cart

200 + 0 = 200 kg m/s

Let's find total mass

Total mass = 40 + 3 = 43 kg

Velocity=Mounmentum/mass

V = P/M

Velocity = 200/43

Hence :-

Velocity of cart is when start moving is 4.65 m/s.

@ananya