Two identical blocks A and B,each of mass m,r | vedclass

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

Question

(A) $1$. When block $C$ collides with block $A$,it sticks to it (assuming perfectly inelastic collision as no other information is given). By conserva

Vedclass · Accepted Answer

$v\sqrt{\frac{m}{2K}}$

Vedclass · Answer

(A) $1$. When block $C$ collides with block $A$,it sticks to it (assuming perfectly inelastic collision as no other information is given). By conservation of linear momentum for the system $(C+A)$: $mv = (m+m)V_A \Rightarrow V_A = v/2$.$2$. Now,we have a system where block $A$ (with $C$ attached) moves with velocity $v/2$ towards block $B$,which is initially at rest.$3$. The maximum compression in the spring occurs when both blocks $(A+C)$ and $B$ move with the same velocity $V_{cm}$ relative to the center of mass frame.$4$. By conservation of linear momentum for the whole system $(C+A+B)$: $mv = (m+m+m)V_{cm} \Rightarrow V_{cm} = v/3$.$5$. By conservation of mechanical energy: $\frac{1}{2}(2m)(v/2)^2 = \frac{1}{2}(3m)(v/3)^2 + \frac{1}{2}Kx_{max}^2$.$6$. $\frac{1}{2}(2m)(v^2/4) = \frac{1}{2}(3m)(v^2/9) + \frac{1}{2}Kx_{max}^2 \Rightarrow \frac{mv^2}{4} = \frac{mv^2}{6} + \frac{1}{2}Kx_{max}^2$.$7$. $\frac{1}{2}Kx_{max}^2 = \frac{mv^2}{4} - \frac{mv^2}{6} = \frac{3mv^2 - 2mv^2}{12} = \frac{mv^2}{12}$.$8$. $x_{max}^2 = \frac{mv^2}{6K} \Rightarrow x_{max} = v\sqrt{\frac{m}{6K}}$.*Note: If the collision is elastic,$C$ comes to rest and $A$ moves with velocity $v$. Then $mv = (m+m)V_{cm} \Rightarrow V_{cm} = v/2$. Energy: $\frac{1}{2}mv^2 = \frac{1}{2}(2m)(v/2)^2 + \frac{1}{2}Kx_{max}^2 \Rightarrow \frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}Kx_{max}^2 \Rightarrow \frac{1}{2}Kx_{max}^2 = \frac{1}{4}mv^2 \Rightarrow x_{max} = v\sqrt{\frac{m}{2K}}$. This matches option $A$.*

Column-$I$	Column-$II$
$(1)$ Work done is zero	$(a)$ By gravitational force (when body moves down)
$(2)$ Work done is positive	$(b)$ Against gravitational force (when body moves up)
$(3)$ Work done is negative	$(c)$ By centripetal force

Similar Questions

$A$ rubber ball is dropped from a height of $5 \ m$ on a planet,where the acceleration due to gravity is not known. On bouncing,it rises to $1.8 \ m$. The ball loses its velocity on bouncing by a factor of

$A$ section of a fixed smooth circular track of radius $R$ in a vertical plane is shown in the figure. $A$ block is released from position $A$ and leaves the track at $B$. The radius of curvature of its trajectory when it just leaves the track at $B$ is:

Match the items in Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Work done is zero $(a)$ By gravitational force (when body moves down)

$(2)$ Work done is positive $(b)$ Against gravitational force (when body moves up)

$(3)$ Work done is negative $(c)$ By centripetal force

Vedclass Test Series

Exam Paper Generator

Online Exam Module