Question

A force separately produces accelerations of 18 ms⁻², 9 ms⁻² and 6 ms⁻² in three bodies of masses P, Q and R respectively. If the same force is applied on a body of mass P+Q+R, then the acceleration of that body is

• A. 3 ms⁻²
• B. 6 ms⁻²
• C. 2 ms⁻²
• D. 33 ms⁻²

Hint:

For problems involving the same force acting on different masses, remember the inverse relationship between mass and acceleration (\(a \propto 1/m\)). The combined body will have a larger mass, so its acceleration must be smaller than the smallest individual acceleration (which was 6 ms⁻²). This can help eliminate some options quickly.

Answer 1

The Correct Option is A

Step 1: Express Mass in terms of Force and Acceleration: From Newton's Second Law, \( F = ma \implies m = \frac{F}{a} \). Let the constant force be \( F \). Mass of body P: \( m_P = \frac{F}{18} \) Mass of body Q: \( m_Q = \frac{F}{9} \) Mass of body R: \( m_R = \frac{F}{6} \)
Step 2: Determine Total Mass: The new body consists of masses P, Q, and R combined. \[ M_{total} = m_P + m_Q + m_R \] \[ M_{total} = \frac{F}{18} + \frac{F}{9} + \frac{F}{6} \] Factor out \( F \): \[ M_{total} = F \left( \frac{1}{18} + \frac{1}{9} + \frac{1}{6} \right) \] Taking LCM (which is 18): \[ M_{total} = F \left( \frac{1 + 2 + 3}{18} \right) = F \left( \frac{6}{18} \right) = \frac{F}{3} \]
Step 3: Calculate the New Acceleration: Let the new acceleration be \( a_{new} \). \[ a_{new} = \frac{F}{M_{total}} \] \[ a_{new} = \frac{F}{F/3} = 3 \, \text{ms}^{-2} \]