Question

A man weighing 70 kg is riding on a cart of mass 30 kg which moves along a level floor at a speed of 3 ms^-1. If he runs on the cart so that his velocity relative to the cart is 4ms^-1 in the direction opposite to the motion of the cart, the speed of centre of mass of the system is

• A. 0.3ms^-1
• B. 0.5ms^-1
• C. 0.2ms^-1
• D. 0.1ms^-1
• E. zero

Answer 1

The Correct Option is C

The center of mass speed \( v_{\text{cm}} \) of the system is given by the formula: \[ v_{\text{cm}} = \frac{m_{\text{man}} v_{\text{man}} + m_{\text{cart}} v_{\text{cart}}}{m_{\text{man}} + m_{\text{cart}}} \] Where:
\( m_{\text{man}} = 70 \, \text{kg} \),
\( m_{\text{cart}} = 30 \, \text{kg} \),
\( v_{\text{man}} = 3 - 4 = -1 \, \text{ms}^{-1} \) (since the man is running opposite to the cart),
\( v_{\text{cart}} = 3 \, \text{ms}^{-1} \).
Substituting these values into the formula: \[ v_{\text{cm}} = \frac{(70)(-1) + (30)(3)}{70 + 30} = \frac{-70 + 90}{100} = \frac{20}{100} = 0.2 \, \text{ms}^{-1} \] Thus, the speed of the center of mass of the system is 0.2 ms\(^{-1}\).

The correct option is (C) : \(0.2\ ms^{-1}\)

Answer 2

Given: 
Mass of the man, m₁ = 70 kg
Mass of the cart, m₂ = 30 kg
Initial speed of the cart with the man, u = 3 m/s
Velocity of the man relative to the cart = 4 m/s in the opposite direction

So, the man's velocity with respect to ground:
$v_1 = 3 - 4 = -1$ m/s
(Since he is moving opposite to the direction of cart's motion)

Velocity of the cart remains: $v_2 = 3$ m/s

Total momentum of the system:
$P = m_1 v_1 + m_2 v_2 = (70)(-1) + (30)(3) = -70 + 90 = 20$ kg·m/s

Total mass of the system:
$M = m_1 + m_2 = 70 + 30 = 100$ kg

Speed of the center of mass:
$v_{cm} = \dfrac{P}{M} = \dfrac{20}{100} = 0.2$ m/s

Answer: 0.2 m/s