A block of mass M moves on a frictionless hor | vedclass

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

Question

(A) According to the law of conservation of mechanical energy,the initial kinetic energy of the block is converted into the elastic potential energy o

Vedclass · Accepted Answer

$\sqrt{Mk} \, L$

Vedclass · Answer

(A) According to the law of conservation of mechanical energy,the initial kinetic energy of the block is converted into the elastic potential energy of the spring at the point of maximum compression.Let $v$ be the velocity of the block just before it hits the spring.$\frac{1}{2} M v^2 = \frac{1}{2} k L^2$Solving for $v$:$v^2 = \frac{k L^2}{M} \implies v = L \sqrt{\frac{k}{M}}$The momentum $p$ of the block is given by $p = Mv$.Substituting the value of $v$:$p = M \left( L \sqrt{\frac{k}{M}} \right) = L \sqrt{M^2 \cdot \frac{k}{M}} = L \sqrt{Mk} = \sqrt{Mk} \, L$Thus,the maximum momentum of the block is $\sqrt{Mk} \, L$.

$A$ block of mass $M$ moves on a frictionless horizontal surface and collides with a spring of spring constant $k$. The spring is compressed by a length $L$. What is the maximum momentum of the block after the collision?

Similar Questions

$A$ mechanical system consists of two springs with stiffness coefficients $k_1$ and $k_2$ connected in series. The minimum work to be performed on the system to stretch it by a total displacement $\Delta$ is

$A$ spring with spring constant $k$ when stretched through $1 \, cm$,the potential energy is $U$. If it is stretched by $4 \, cm$,the potential energy will be (in $U$)

An elastic string of unstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. The work done in the second stretching is

The potential energy of a spring...

$A$ sphere of mass $m$ is attached to a spring of spring constant $k$ and is held in an unstretched position over an inclined plane as shown in the figure. After letting the sphere go,find the maximum length by which the spring extends,given the sphere only rolls.

Vedclass Test Series

Exam Paper Generator

Online Exam Module