Question

A body of mass 1 kg is suspended from a spring of force constant \( 600 \, \text{N m}^{-1} \). Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of \( 3 \, \text{m s}^{-1} \) and embedded in it. The amplitude of motion is

• A. 5 cm
• B. 15 cm
• C. 10 cm
• D. 8 cm

Hint:

1. Conservation of momentum for inelastic collision: \(m_1u_1 + m_2u_2 = (m_1+m_2)v_{final}\). 2. New equilibrium position of spring-mass system: \(k x_{eq} = M_{total}g\). 3. For SHM, total energy \( E = \frac{1}{2} M_{total} v^2 + \frac{1}{2} k y^2 = \frac{1}{2} k A^2 \), where \(y\) is displacement from equilibrium and \(v\) is velocity at that displacement. 4. Alternatively, \( v^2 = \omega^2(A^2-y^2) \), where \( \omega = \sqrt{k/M_{total}} \).

Answer 1

The Correct Option is A

Let \( M = 1 \) kg (suspended mass), \( k = 600 \, \text{N m}^{-1} \).
Let \( m = 0.
5 \) kg (hitting mass), \( v_0 = 3 \, \text{m s}^{-1} \) (its initial velocity).
Initially, mass M is suspended, so the spring is stretched by \( x_0 \) such that \( Mg = kx_0 \).
\( x_0 = \frac{Mg}{k} = \frac{1 \times 10}{600} = \frac{1}{60} \) m (assuming \( g=10 \, \text{m/s}^2 \)).
This is the initial equilibrium position.
Collision: The mass m hits M and embeds.
This is a perfectly inelastic collision.
Let \( v \) be the velocity of the combined mass \( (M+m) \) just after collision.
By conservation of linear momentum (assuming collision time is very short, spring force impulse is negligible): \( mv_0 + M(0) = (M+m)v \) (Initial velocity of M is 0 as it's suspended at rest).
\( 0.
5 \times 3 = (1+0.
5)v \) \( 1.
5 = 1.
5v \implies v = 1 \, \text{m s}^{-1} \) upwards.
New system: Combined mass \( M' = M+m = 1+0.
5 = 1.
5 \) kg.
Spring constant \( k = 600 \, \text{N m}^{-1} \).
The new equilibrium position \( x_{eq}' \) for mass \( M' \) is where \( M'g = kx_{eq}' \).
\( x_{eq}' = \frac{M'g}{k} = \frac{1.
5 \times 10}{600} = \frac{15}{600} = \frac{1}{40} \) m.
This new equilibrium position is measured from the unstretched length of the spring.
At the moment of collision, the system (mass M) was at its equilibrium position \(x_0 = 1/60\) m.
The collision occurs at this position.
So, just after collision, the combined mass \(M'\) is at \(x_0 = 1/60\) m from the unstretched position, and has velocity \( v = 1 \, \text{m s}^{-1} \) upwards.
The SHM occurs about the new equilibrium position \( x_{eq}' = 1/40 \) m.
The displacement of the combined mass from its new equilibrium position at \( t=0 \) (just after collision) is: \( y = x_0 - x_{eq}' = \frac{1}{60} - \frac{1}{40} = \frac{2-3}{120} = -\frac{1}{120} \) m.
The negative sign means it is \(1/120\) m above the new equilibrium position.
The velocity at this displacement is \( v = 1 \, \text{m s}^{-1} \).
For SHM, the total energy is constant: \( E = \frac{1}{2}kA^2 \), where A is the amplitude.
Also, \( E = \frac{1}{2}M'v_y^2 + \frac{1}{2}ky^2 \), where \(y\) is displacement from new equilibrium.
Angular frequency of new SHM: \( \omega' = \sqrt{\frac{k}{M'}} = \sqrt{\frac{600}{1.
5}} = \sqrt{\frac{600}{3/2}} = \sqrt{400} = 20 \, \text{rad/s} \).
The velocity \(v\) and displacement \(y\) from equilibrium are related to amplitude \(A\) by \( v^2 = \omega'^2 (A^2 - y^2) \).
\( (1)^2 = (20)^2 \left( A^2 - \left(-\frac{1}{120}\right)^2 \right) \) \( 1 = 400 \left( A^2 - \frac{1}{14400} \right) \) \( \frac{1}{400} = A^2 - \frac{1}{14400} \) \[ A^2 = \frac{1}{400} + \frac{1}{14400} = \frac{36}{14400} + \frac{1}{14400} = \frac{37}{14400} \] \[ A = \sqrt{\frac{37}{14400}} = \frac{\sqrt{37}}{120} \, \text{m} \] \( \sqrt{36}=6, \sqrt{49}=7 \).
\( \sqrt{37} \approx 6.
08 \).
\( A \approx \frac{6.
08}{120} \approx \frac{6}{120} = \frac{1}{20} = 0.
05 \, \text{m} = 5 \, \text{cm} \).
More precisely: \( \frac{\sqrt{37}}{120} \text{ m} \approx \frac{6.
08276}{120} \text{ m} \approx 0.
050689 \text{ m} \approx 5.
07 \text{ cm} \).
This is very close to 5 cm.
Let's verify the use of \(g=10 \text{ m/s}^2 \).
If \(g=9.
8 \text{ m/s}^2\): \(x_0 = 1 \times 9.
8 / 600 = 9.
8/600 \).
\(x_{eq}' = 1.
5 \times 9.
8 / 600 = 14.
7/600 \).
\(y = x_0 - x_{eq}' = (9.
8-14.
7)/600 = -4.
9/600 \) m.
\(v=1 \text{ m/s}\).
\( \omega' = 20 \text{ rad/s}\).
\( 1^2 = 20^2 (A^2 - (-4.
9/600)^2) \) \( A^2 = \frac{1}{400} + \left(\frac{4.
9}{600}\right)^2 = \frac{1}{400} + \frac{24.
01}{360000} = \frac{900+24.
01}{360000} = \frac{924.
01}{360000} \).
\( A = \frac{\sqrt{924.
01}}{600} \approx \frac{30.
397}{600} \approx 0.
05066 \) m \( \approx 5.
07 \) cm.
The result is consistently around 5 cm.
So option (1) is correct.