Medium
$A$ body of mass $m_1$ moving with an unknown velocity $v_1 \hat{i}$ undergoes a one-dimensional collision with a body of mass $m_2$ moving with velocity $v_2 \hat{i}$. After the collision,the bodies $m_1$ and $m_2$ move with velocities $v_3 \hat{i}$ and $v_4 \hat{i}$ respectively. If $m_2 = 0.5\, m_1$ and $v_3 = 0.5\, v_1$,find $v_1$.
- A$v_4 - \frac{v_2}{2}$
- B$v_4 - \frac{v_2}{4}$
- C$v_4 - v_2$
- D$v_4 + v_2$
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Two particles,of masses $M$ and $2M$,moving as shown,with speeds of $10\, m/s$ and $5\, m/s$,collide elastically at the origin. After the collision,they move along the indicated directions with speeds $v_1$ and $v_2$ respectively. The values of $v_1$ and $v_2$ are nearly:
DifficultJEE MAIN 2019
View SolutionIn the figure shown,two identical balls of mass $M$ and radius $R$ each are placed in contact with each other on a frictionless horizontal surface. $A$ third ball of mass $M$ and radius $R$ is coming down vertically and has a velocity $v_0$ when it simultaneously hits the two balls and comes to rest. Then,each of the two bigger balls will move after the collision with a speed equal to:
Which of the following graphs correctly represents the relationship between kinetic energy $(E)$,potential energy $(U)$,and height $(h)$ from the ground for a particle moving vertically?
Easy
View Solution$A$ block of mass $5 \ kg$ is lifted to a height of $5 \ m$ by a force of $60 \ N$. Find:
$(1)$ Work done by the force.
$(2)$ Potential energy of the block at $5 \ m$.
$(3)$ Kinetic energy of the block at $5 \ m$.
$(4)$ Velocity of the block at $5 \ m$. (Take $g = 9.8 \ m/s^2$)
$(1)$ Work done by the force.
$(2)$ Potential energy of the block at $5 \ m$.
$(3)$ Kinetic energy of the block at $5 \ m$.
$(4)$ Velocity of the block at $5 \ m$. (Take $g = 9.8 \ m/s^2$)
Medium
View Solution$A$ particle of unit mass is moving along the $x$-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column $I$ ($a$ and $U_0$ are constants). Match the potential energies in column $I$ to the corresponding statement$(s)$ in column $II$.
Column $I$
Column $II$
$(A) U_1(x) = \frac{U_0}{2} \left[1 - \left(\frac{x}{a}\right)^2\right]^2$
$(P)$ The force acting on the particle is zero at $x = a$.
$(B) U_2(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2$
$(Q)$ The force acting on the particle is zero at $x = 0$.
$(C) U_3(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2 \exp \left[-\left(\frac{x}{a}\right)^2\right]$
$(R)$ The force acting on the particle is zero at $x = -a$.
$(D) U_4(x) = \frac{U_0}{2} \left[\frac{x}{a} - \frac{1}{3}\left(\frac{x}{a}\right)^3\right]$
$(S)$ The particle experiences an attractive force towards $x = 0$ in the region $|x| < a$.
$(T)$ The particle with total energy $\frac{U_0}{4}$ can oscillate about the point $x = -a$.
| Column $I$ | Column $II$ |
| $(A) U_1(x) = \frac{U_0}{2} \left[1 - \left(\frac{x}{a}\right)^2\right]^2$ | $(P)$ The force acting on the particle is zero at $x = a$. |
| $(B) U_2(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2$ | $(Q)$ The force acting on the particle is zero at $x = 0$. |
| $(C) U_3(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2 \exp \left[-\left(\frac{x}{a}\right)^2\right]$ | $(R)$ The force acting on the particle is zero at $x = -a$. |
| $(D) U_4(x) = \frac{U_0}{2} \left[\frac{x}{a} - \frac{1}{3}\left(\frac{x}{a}\right)^3\right]$ | $(S)$ The particle experiences an attractive force towards $x = 0$ in the region $|x| < a$. |
| $(T)$ The particle with total energy $\frac{U_0}{4}$ can oscillate about the point $x = -a$. |
DifficultIIT 2015
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