Question

girl of mass 40 kg jumps with the horizontal velocity of 5ms-1 onto a stationary card with fictional Les wheel the mass of the Kartus 3kd what is a velocity as the card start moving assume that there is no external unbalanced for working in the horizontal direction ​

Answer 1

Answer:

Given :   Mgirl=40 kg           ugirl=5 m/s        Mcart=3 kg

Initial velocity of cart  ucart=0

Let the final velocity of cart be  V.

Applying conservation of linear momentum :    Pi=Pf

Or    Mgirlugirl+Mcartucart=(Mgirl+Mcart)V

Or    40×5+3×0=(40+3)V

⟹ V=43200 m/s

Answer 2

Given :

Condition Given : A girl of a certain mass jumps horizontally with a velocity on a resting cart of given mass with friction less wheels. When Girl jumps on the Cart, it will start moving along with the Girl.

• Mass of Girl, m₁ = 40 kg
• Initial velocity of Girl, u₁ = 5 ms⁻¹

• Mass of Cart, m₂ = 3 kg
• Initial velocity of Cart, u₂ = 0

To Find :

• Final velocity of Girl = Final velocity of Cart = v

Knowledge required :

• Law of conservation of momentum

For two or more bodies in an isolated system acting upon each other, the total momentum of the system remains constant unless an external force is applied. therefore,

→ Initial momentum of system = Final momentum of system

→ m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

[ where m₁ and m₂ are masses, u₁ and u₂ are initial velocities, v₁ and v₂ are Final velocities of two bodies respectively ]

Solution :

Using Law of conservation of momentum

→ m₁u₁ + m₂u₂ = m₁v + m₂v

→ ( 40 ) ( 5 ) + ( 3 ) ( 0 ) = v ( m₁ + m₂ )

→ 200 + 0 = v ( 40 + 3 )

→ 200 = 43  v

→ v = 4.65 ms⁻¹

Therefore,

• Cart along with Girl will move with a velocity of 4.65 ms⁻¹ in the Initial direction of motion of Girl.