A 1,kg mass is attached to a spring of force | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Applying the principle of conservation of linear momentum during the perfectly inelastic collision:$m_2 u = (m_1 + m_2) v$$0.5 	imes 3 = (1 + 0.5

Vedclass · Accepted Answer

$5\,cm, \frac{\pi}{10}\,s$

Vedclass · Answer

(A) Applying the principle of conservation of linear momentum during the perfectly inelastic collision:$m_2 u = (m_1 + m_2) v$$0.5 	imes 3 = (1 + 0.5) v$$1.5 = 1.5 v \Rightarrow v = 1\,m/s$After the collision,the combined mass $(M = m_1 + m_2 = 1.5\,kg)$ oscillates with the spring. Using the conservation of mechanical energy:$\frac{1}{2} M v^2 = \frac{1}{2} k A^2$$A = v \sqrt{\frac{M}{k}} = 1 	imes \sqrt{\frac{1.5}{600}} = \sqrt{\frac{1}{400}} = \frac{1}{20}\,m = 0.05\,m = 5\,cm$The time period of oscillation is given by:$T = 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{1.5}{600}} = 2\pi \sqrt{\frac{1}{400}} = 2\pi 	imes \frac{1}{20} = \frac{\pi}{10}\,s$Thus,the amplitude is $5\,cm$ and the time period is $\frac{\pi}{10}\,s$.

Similar Questions

Two particles are in $SHM$ in a straight line. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0$,one particle is at displacement $y_1 = +A$ and the other at $y_2 = -A/2$,and they are approaching towards each other. After what time do they cross each other?

$A$ mass $m$ oscillates with simple harmonic motion with frequency $f = \frac{\omega}{2\pi}$ and amplitude $A$ on a spring with constant $K$. Therefore:

$A$ particle executing simple harmonic motion has an amplitude of $6\, cm$. Its acceleration at a distance of $2 \,cm$ from the mean position is $8\, cm/s^2$. The maximum speed of the particle is ... $cm/s$.

Two oscillating systems,a simple pendulum and a vertical spring-mass system,have the same time period of motion on the surface of the Earth. If both are taken to the moon,then-

Vedclass Test Series

Exam Paper Generator

Online Exam Module