A cart and a man move towards each other. The masses of cart and man are 25 kg and 50 kg respectively. The velocity of cart is 0.2 m/s. When the man approaches the cart, he jumps into it. After this the cart carrying the man acquires a velocity of 0.6 m/s in the direction in which the man was walking. What is the velocity with which the man was walking?
Solve by using law of conservation of momentum
Answers
Given :-
→ Mass of the cart (m₁) = 25 kg
→ Mass of the man (m₂) = 50 kg
→ Velocity of the cart (u₁) = 0.2 m/s
→ Combined velocity of the man and the cart (v) = 0.6 m/s
To find :-
→ Velocity with which the man was walking (u₂) .
Solution :-
In the question, it is given that the man and the cart are moving towards each other. This means that they are moving in opposite directions and we have to take any one of the velocites as negative.
∴ Velocity of the cart = - 0.2 m/s
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According to Law of Conservation of momentum, we have :-
m₁u₁ + m₂u₂ = (m₁ + m₂)v
Substituting values, we get :-
⇒ 25(-0.2) + 50(u₂) = (50 + 25)0.6
⇒ -5 + 50u₂ = 75(0.6)
⇒ -5 + 50u₂ = 45
⇒ 50u₂ = 50
⇒ u₂ = 50/50
⇒ u₂ = 1 m/s
Thus, the man was walking with a velocity of 1 m/s .
Given :-
A cart and a man move towards each other. The masses of cart and man are 25 kg and 50 kg respectively. The velocity of cart is 0.2 m/s. When the man approaches the cart, he jumps into it. After this the cart carrying the man acquires a velocity of 0.6 m/s in the direction in which the man was walking.
To Find :-
Velocity with which the man was walking?
Solution :-
We know that
Where
m₁ = Mass of the first object(cart)
u₁ = Initial Velocity of first object(cart)
m₂ = Mass of second object(Man)
u₂ = Initial Velocity of second object(Man)
v₁ = Final velocity of first object(Cart)
v₂ = Final velocity of second object(Man)