(A) Let the mass of the incoming object be $m$ and its velocity be $v$. The kinetic energy is $E = \frac{1}{2}mv^2$ and momentum is $P = mv$. Thus,$m

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First increases then decreases with linear momentum

(A) Let the mass of the incoming object be $m$ and its velocity be $v$. The kinetic energy is $E = \frac{1}{2}mv^2$ and momentum is $P = mv$. Thus,$m = \frac{P^2}{2E}$.For a head-on elastic collision with a stationary mass $M$,the kinetic energy transferred to $M$ is given by $\Delta K = \frac{4mM}{(m+M)^2} E$.Substituting $m = \frac{P^2}{2E}$ into the expression:$\Delta K = \frac{4(\frac{P^2}{2E})M}{(\frac{P^2}{2E} + M)^2} E = \frac{2P^2 M}{(\frac{P^2 + 2ME}{2E})^2} E = \frac{8ME P^2}{(P^2 + 2ME)^2}$.Let $f(P) = \frac{8ME P^2}{(P^2 + 2ME)^2}$. As $P \to 0$,$f(P) \to 0$. As $P \to \infty$,$f(P) \to 0$. Since $f(P) > 0$ for $P > 0$,the function must increase to a maximum and then decrease. Therefore,the energy transferred first increases and then decreases with linear momentum $P$.

Class 11 Physics Work, Energy, Power and Collision Mix Examples-Work, Energy, Power and Collision

Difficult

An object of mass $M$ is at rest on a smooth horizontal surface. Objects of different masses collide head-on elastically with the object of mass $M$. All colliding objects have the same fixed kinetic energy $E$,and in each case,mass $M$ is initially at rest. The kinetic energy transferred to the stationary mass $M$ depends on the linear momentum $P$ of the incoming colliding mass. How does the energy transferred to $M$ vary with the linear momentum $P$?

A
First increases then decreases with linear momentum
B
First decreases then increases with linear momentum
C
Always decreases with linear momentum
D
Always increases with linear momentum

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Work, Energy, Power and Collision

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Mix Examples-Work, Energy, Power and Collision

Two beads $A$ and $B$ of masses $m_1$ and $m_2$ respectively are placed on a smooth vertical circular wire of radius $R$ as shown in the figure. Bead $A$ is given a very gentle push so that it slides down and collides with bead $B$ and comes to rest. After the collision,bead $B$ reaches the height of the center of the circle. Then $m_1 : m_2$ is equal to:

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$A$ bullet of mass $0.02 \ kg$ travelling horizontally with velocity $250 \ ms^{-1}$ strikes a block of wood of mass $0.23 \ kg$ which rests on a rough horizontal surface. After the impact,the block and bullet move together and come to rest after travelling a distance of $40 \ m$. The coefficient of sliding friction of the rough surface is $\left(g=9.8 \ ms^{-2}\right)$

DifficultAP EAMCET 2009

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Two blocks $A$ and $B$,each of mass $m$,are connected by a massless spring of natural length $L$ and spring constant $K$. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in the figure. $A$ third identical block $C$,also of mass $m$,moves on the floor with a speed $v$ along the line joining $A$ and $B$ and collides with $A$. Then:

DifficultIIT 1993

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$A$ shell is fired from a cannon with velocity $v \text{ m/s}$ at an angle $\theta$ with the horizontal direction. At the highest point in its path,it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon. The speed in $\text{m/s}$ of the other piece immediately after the explosion is:

DifficultIIT 1984

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Two statements are given below. Select the option that correctly explains both statements.
Statement-$1$: In a perfectly elastic collision between two particles moving in the same direction,they do not lose all their energy.
Statement-$2$: The principle of conservation of momentum is valid for all types of collisions.

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