$A$ particle of mass $M$ moving along the $X$-axis with speed $V_0$ collides with and sticks to another particle of mass $m$ moving along the $Y$-axis with speed $V_0$. What will be the velocity of the combined mass after the collision?

  • A
    $\frac{M\hat{i} + m\hat{j}}{M + m} V_0$
  • B
    $\frac{m\hat{i} + M\hat{j}}{M + m} V_0$
  • C
    $(m\hat{i} + M\hat{j}) V_0$
  • D
    $\frac{M\hat{i} + m\hat{j}}{M} V_0$

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Three blocks $A, B$ and $C$ are kept as shown in the figure. The coefficient of friction between $A$ and $B$ is $0.2$,$B$ and $C$ is $0.1$,and $C$ and the ground is $0.0$. The masses of $A, B$ and $C$ are $3\, kg, 2\, kg$ and $1\, kg$ respectively. $A$ is given a horizontal velocity of $10\, m/s$. Blocks $A, B$ and $C$ always remain in contact and move together as a single system. The total work done by friction will be ........ $J$.

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$A$ point mass of $1 \, kg$ collides elastically with a stationary point mass of $5 \, kg$. After their collision, the $1 \, kg$ mass reverses its direction and moves with a speed of $2 \, m/s$. Which of the following statement(s) is (are) correct for the system of these two masses?
$(A)$ Total momentum of the system is $3 \, kg \cdot m/s$
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$(C)$ Kinetic energy of the centre of mass is $0.75 \, J$
$(D)$ Total kinetic energy of the system is $4 \, J$

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