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

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

$A$ system of two identical rods ($L$-shaped) of mass $m$ and length $l$ are resting on a peg $P$ as shown in the figure. If the system is displaced in its plane by a small angle $\theta$,find the period of oscillations:

$A$ particle is executing $SHM$ with amplitude $A,$ time period $T,$ maximum acceleration $a_0$ and maximum velocity $v_0.$ It starts from the mean position at $t=0.$ At time $t,$ it has displacement $A/2,$ acceleration $a,$ and velocity $v.$ Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module