Question

A girl of mass $40\,{\text{kg}}$ jumps with a horizontal velocity of $5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ onto a stationary cart with frictionless wheels. The mass of the cart is $3\,{\text{kg}}$. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.

Answer 1

Use the formula for the linear momentum of an object. Also, use the formula for law of conservation of linear momentum for inelastic collision between the two object. Determine the velocity of the cart before the girl jumps on it. Rewrite this formula for inelastic collision between the girl and the cart and calculate the velocity of the girl-cart system after the girl jumps on the cart.

Formulae used:
The linear momentum $P$ of an object is given by
$P = mv$ …… (1)
Here, $m$ is the mass of the object and $v$ is the velocity of the object.
The expression for the law of conservation of linear momentum for inelastic collision is given by,
${m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)V$ …... (2)
Here, ${m_1}$ and ${m_2}$ are the masses of the two objects, ${v_1}$ and ${v_2}$ are the velocities of the two objects before collision and $V$ is the velocity of the system of two object after collision.

Complete step by step answer:
We have given that the mass of the girl is $40\,{\text{kg}}$ and the mass of the cart is $3\,{\text{kg}}$.
${m_{girl}} = 40\,{\text{kg}}$
$\Rightarrow{m_{cart}} = 3\,{\text{kg}}$
We have given that the horizontal velocity of the girl before jumping on the cart is $5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.
${v_{girl}} = 5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
Initially, the cart is at rest. Hence, the velocity of the cart before the girl jumps on it is $0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.
${v_{cart}} = 0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
We have asked to determine the velocity of the cart after the girl jumps on it which is the velocity of the girl-cart system as the girl jumps on the cart and the cart starts moving.The collision between the girl and the cart is inelastic.

According to the law of conservation of the linear momentum, the initial linear momentum of the girl and the cart is equal to the linear momentum of the girl-cart system.Hence, the equation (2) for the law of conservation of the linear momentum for inelastic collision becomes
${m_{girl}}{v_{girl}} + {m_{cart}}{v_{cart}} = \left( {{m_{girl}} + {m_{cart}}} \right)V$
Substitute $40\,{\text{kg}}$ for ${m_{girl}}$, $3\,{\text{kg}}$ for ${m_{cart}}$, $5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ for ${v_{girl}}$ and $0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ for ${v_{cart}}$ in the above equation.
$\left( {40\,{\text{kg}}} \right)\left( {5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right) + \left( {3\,{\text{kg}}} \right)\left( {0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right) = \left( {40\,{\text{kg}} + 3\,{\text{kg}}} \right)V$
$\Rightarrow 200 + 0 = \left( {43} \right)V$
$\Rightarrow V = \dfrac{{200}}{{43}}$
$\therefore V = 4.65\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$

Hence, the velocity of the cart after the girl jumps on it is $4.65\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.

Note: The students may get confused as we have asked to calculate the velocity of the cart as it starts moving. But we know that the cart is initially at rest and it starts moving when the girl jumps on it and the velocity of the cart after the girl jumps on it is the same as the velocity of the girl-cart system as the girl is on the cart after jumping. Hence, the velocity of the girl-cart system is the required velocity of the cart.