A 1, kg block moves towards a light spring wi | vedclass

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

Question

(B) Given: Mass $m = 1\, kg$,initial velocity $v_1 = 8\, m/s$,compression $x = 3\, m$.Initial momentum $P_1 = m v_1 = 1 	imes 8 = 8\, kg\cdot m/s$.Fi

Vedclass · Accepted Answer

$16/3\, N/m$

Vedclass · Answer

(B) Given: Mass $m = 1\, kg$,initial velocity $v_1 = 8\, m/s$,compression $x = 3\, m$.Initial momentum $P_1 = m v_1 = 1 	imes 8 = 8\, kg\cdot m/s$.Final momentum $P_2 = P_1 / 2 = 4\, kg\cdot m/s$.Since $P_2 = m v_2$,we have $v_2 = P_2 / m = 4 / 1 = 4\, m/s$.By the law of conservation of energy,the loss in kinetic energy is equal to the potential energy stored in the spring:$\frac{1}{2} m v_1^2 - \frac{1}{2} m v_2^2 = \frac{1}{2} k x^2$$m(v_1^2 - v_2^2) = k x^2$$1 	imes (8^2 - 4^2) = k 	imes 3^2$$1 	imes (64 - 16) = 9k$$48 = 9k$$k = 48 / 9 = 16 / 3\, N/m$.

$A$ $1\, kg$ block moves towards a light spring with a velocity of $8\, m/s$. When the spring is compressed by $3\, m$,its momentum becomes half of the original momentum. The spring constant of the spring is:

Similar Questions

When a long spring of $4 \ cm$ is stretched by $1 \ cm$,the potential energy stored in the spring is $U$. If it is stretched by $4 \ cm$,the potential energy stored in it is (in $U$)

$A$ spring of force constant $800\, N/m$ has an extension of $5\, cm$. The work done in extending it from $5\, cm$ to $15\, cm$ is ............. $J$.

The potential energy of a spring...

$A$ spring with a force constant of $100 \ N/m$ is stretched by $5 \ cm$. Calculate the work done. (in $J$)

The work done in stretching a spring of force constant $5 \times 10^3 \ N/m$ from $5 \ cm$ to $10 \ cm$ is ...... $N-m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module