Question

A man and a cart move towards each other. The man weighs 64 kg and the cart 32 kg. The velocity of the man is $5.4\,kg/hr$and that of the cart is $1.8\,km/hr$. When the man approaches the cart, he jumps on it. What will be the velocity of the cart carrying the man?
A. $30\,km/hr$
B. $3\,km/hr$
C. $1.8\,km/hr$
D. Zero

Answer 1

The initial linear momentum of the system of a man and a cart remains the same after the man jumps on the cart. Since the man jumps on the cart, the final velocity of the man and the cart is the same.

Complete step by step answer:
The linear momentum of a body of mass $m$ moving with velocity $v$ is the product of its mass and velocity.
$P = mv$

According to the law of conservation of linear momentum, the total linear momentum of the system remains conserved before and after the collision.

In this question, the system consists of a man and a cart moving towards each other. After the man jumps on the cart, the momentum of the man is transferred to the cart.

Suppose the mass of the man is ${m_m}$ and the mass of the cart is ${m_c}$. The initial velocity of the man is ${u_m}$and the initial velocity of the cart is ${u_c}$ and the final velocity of the man after he jumps on the cart is ${v_m}$ and the final velocity of the cart is ${v_c}$.

Let us apply the law of conservation of linear momentum to the system of a man and a cart as follows,
${m_m}{u_m} + {m_c}{u_c} = {m_m}{v_m} + {m_c}{v_c}$

Suppose the man is moving along the positive x direction and the cart is moving along negative x direction. Therefore, we have to substitute $- {u_c}$ for ${u_c}$ in the above equation.
${m_m}{u_m} - {m_c}{u_c} = {m_m}{v_m} + {m_c}{v_c}$

Since the man jumps on the cart, the velocity of the man is the same as the velocity of the cart. Therefore, we can substitute v for ${v_m}$ and ${v_c}$ in the above equation.
${m_m}{u_m} - {m_c}{u_c} = {m_m}v + {m_c}v$
$\Rightarrow {m_m}{u_m} - {m_c}{u_c} = \left( {{m_m} + {m_c}} \right)v$

Substitute 64 kg for ${m_m}$, $5.4\,km/hr$ for ${u_m}$, 32 kg for ${m_c}$, $1.8\,km/hr$ for ${u_c}$ in the above equation.
$\left( {64\,kg} \right)\left( {5.4\,km/hr} \right) - \left( {32\,kg} \right)\left( {1.8\,km/hr} \right) = \left( {64\,kg + 32\,kg} \right)v$
$\Rightarrow 345.6\,kg\,km/hr - 57.6\,kg\,km/hr = \left( {96\,kg} \right)v$
$\Rightarrow 288\,km/hr = 96\,v$
$\Rightarrow v = \dfrac{{288\,km/hr}}{{96}}$
$\therefore v = 3\,km/hr$

Since the velocity of the cart is positive, the cart has started to move along the positive x direction which is the direction of the man.

So, the correct answer is “Option B”.

Note:
While solving such problems, the direction of the body moving is important. One has to set the direction of the moving body. In this problem, the direction of a man and a cart is opposite to each other. Therefore, the sign of their respective velocities should be opposite.