Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) s

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5.Work, Energy, Power and Collision

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Two identical uniform discs roll without slipping on two different surfaces $AB$ and $CD$ (see figure) starting at $A$ and $C$ with linear speeds $v _1$ and $v _2$, respectively, and always remain in contact with the surfaces. If they reach $B$ and $D$ with the same linear speed and $v_1=3 \ m / s$, then $v_2$ in $m / s$ is $\left(g=10 \ m / s ^2\right)$

$5$

$6$

$7$

$8$

(IIT-2015)

Solution

Kinetic energy of a pure rolling disc having velocity of centre of mass $v=\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{m R^2}{2}\right) \frac{v^2}{R^2}=\frac{3}{4} m v^2$

So, $\frac{3}{4} m(3)^2+m g(30)=\frac{3}{4} m\left(v_2\right)^2+m g(27) \quad \therefore v_2=7 m / s$

Standard 11

Physics

At time $t=0$ is particle starts moving along the $x-$axis. If its kinetic energy increases uniformly with time $t$, the net force acting on it must be proportional to

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(AIEEE-2011)

A bomb of $12\, kg$ explodes into two pieces of masses $4\, kg$ and $8\, kg$. The velocity of $8\, kg$ mass is $6\, m/sec$. The kinetic energy of the other mass is ………….. $\mathrm{J}$

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If a body of mass $200\, g$ falls from a height $200 \,m$ and its total $P.E.$ is converted into $K.E.$ at the point of contact of the body with earth surface, then what is the decrease in $P.E.$ of the body at the contact $(g = 10\,m/{s^2})$ ………… $\mathrm{J}$

easy

Two bodies of masses $m$ and $4 \,m$ are moving with equal $K.E.$ The ratio of their linear momentums is

easy

(AIIMS-1999)

A particle $A$ is projected vertically upwards. Another particle $B$ is projected at an angle of $45^{\circ}$. Both reach the same height. The ratio of the initial kinetic energy of $A$ to that of $B$ is

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Solution

Similar Questions

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