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A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

$\frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$

$\frac{1}{3} v \hat{i} + \frac{2}{3} v \hat{j}$

$\frac{2}{3} v \hat{i} + \frac{1}{3} v \hat{j}$

$\frac{3}{2} v \hat{i} + \frac{1}{4} v \hat{j}$

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A circular disk of moment of inertia $I_{t}$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $ω_{i} .$ Another disk of moment of inertia $I_{b}$ is dropped coaxially onto the rotation disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $ω_{f} .$ The energy lost by the initially rotating disc to friction is

$\frac{1}{2} \frac{I_{b}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{t}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{b} - I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{b} I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

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Explanation

Loss of energy, $∆ E = \frac{1}{2} I_{t} ω_{i}^{2} - \frac{1}{2} \frac{I_{t}^{2} ω_{i}^{2}}{(I_{t} + I_{b})}$

= $\frac{1}{2} \frac{I_{b} I_{t} ω_{i}^{2}}{(I_{t} + I_{b})}$

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A man of 50 kg mass is standing in a gravity free space at a height of 10m above the floor. He throws a stone of 0.5 kg mass downwards with a speed $2 m s^{- 1} .$ When the stone reaches the floor, the distance of the man above the floor will be

9.9m

10.1m

10m

20m

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Explanation

mr = constant

$m_{1} r_{1} = m_{2} r_{2}$

$\Rightarrow r_{2} = \frac{m_{1} r_{1}}{m_{2}}$

= $\frac{0.5 \times 10}{50} = 0.1$

The distance of the man above the floor (total height)=10+0.1=10.1m.

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An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of $12 {ms}^{- 1}$ and 2 kg second part moving with a velocity of $8 {ms}^{- 1} .$ If the third part flies off with a velocity of $4 {ms}^{- 1},$ its mass would be

$5 kg$

$7 kg$

$17 kg$

$3 kg$

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If $\vec{F}$ is the force acting on a particle having position vector $\vec{r}$ and $\vec{τ}$ be the torque of this force about the origin, then

$\vec{r} . \vec{τ} \neq 0 and \vec{F} . \vec{τ} = 0$

$\vec{r} . \vec{τ} > 0 and \vec{F} . \vec{τ} < 0$

$\vec{r} . \vec{τ} = 0 and \vec{F} . \vec{τ} = 0$

$\vec{r} . \vec{τ} = 0 and \vec{F} . \vec{τ} \neq 0$

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Explanation

Key Idea Torque is an axial vector ie, its direction is always perpendicular to the plane containing vectors $\vec{r}$ and $\vec{F}$ .

$\vec{τ} = \vec{r} \times \vec{F}$

Torque is perpendicular to both $\vec{r}$ and $\vec{F}$

∴ $\vec{τ} \cdot \vec{r} = 0$ $\vec{F} \cdot \vec{τ} = 0$

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A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $ω .$ If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity -

$\frac{ω (M - 2 m)}{M + 2 m}$

$\frac{ω M}{M + 2 m}$

$\frac{ω (M + 2 m)}{M}$

$\frac{ω M}{M + m}$

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Explanation

Key Idea

Apply law of conservation of angular momentum.

$I_{1} ω_{1} = I_{2} ω_{2}$

In the given case

$I_{1} = M R^{2}$

$I_{2} = M R^{2} + 2 m R^{2}$

$ω_{1} = ω$

Then $ω_{2} = \frac{I_{1}}{I_{2}} ω = \frac{M}{M + 2 m} ω$

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Two bodies of mass 1kg and 3kg have position vectors $\hat{i} + 2 \hat{j} + \hat{k}$ and $- 3 \hat{i} - 2 \hat{j} + \hat{k}$ , respectively. The centre of mass of this system has a position vector -

$- 2 \hat{i} + 2 \hat{k}$

$- 2 \hat{i} - \hat{j} + \hat{k}$

$2 \hat{i} - \hat{j} - 2 \hat{k}$

$- \hat{i} + \hat{j} + \hat{k}$

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Explanation

The position vector of centre of mass

$\vec{r} = \frac{m_{1} \vec{r_{1}} + m_{2} \vec{r_{2}}}{m_{1} + m_{2}}$

= $\frac{1 (\hat{i} + 2 \hat{j} + \hat{k}) + 3 (- 3 \hat{i} - 2 \hat{j} + \hat{k})}{1 + 3}$

= $\frac{1}{4} (- 8 \hat{i} - 4 \hat{j} + 4 \hat{k})$

$= - 2 \hat{i} - \hat{j} + \hat{k}$

The centre of mass changes its position only under the translatory motion. There is no effect of rotatory motion on centre of mass of the body.

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Four identical thin rods each of mass M and length t, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

$\frac{4}{3} M t^{2}$

$\frac{2}{3} M t^{2}$

$\frac{13}{3} M t^{2}$

$\frac{1}{3} M t^{2}$

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Explanation

Apply theorem of parallel axisand the total moment of interia will be the sum of moment of inertia of each rod.

Moment of inertia of rod about an axis through its centre of mass and perpendicular to rod + (mass of rod) $\times$ (perpendicular distance between two axes)

= $\frac{M t^{2}}{12} + M {(\frac{t}{2})}^{2} = \frac{M t^{2}}{3}$

Moment of inertia of the system = $\frac{M t^{2}}{3} \times 4$

= $\frac{4}{3} M t^{2}$

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A shell of mass 200g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is -

100 $m s^{- 1}$

80 $m s^{- 1}$

40 $m s^{- 1}$

120 $m s^{- 1}$

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Explanation

In the given problem conservation of linear of momentum yields.

$m_{1} v_{1} + m_{2} v_{2} = 0$

or $4 v_{1} + 0.2 v_{2} = 0 . . . . (i)$

Conservation of energy yields.

$\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = 1050 0 r \frac{1}{2} \times 4 v_{1}^{2} + \frac{1}{2} \times 0.2 \times v_{2}^{2} = 1050 o r 2 v_{1}^{2} + 0.1 v_{2}^{2} = 1050 . . . . (i i)$

Solving Eqs. (i) and (ii) , we have

$v_{2} = 100 m / s$

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A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is $90^{°}$ . The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

$\frac{M L^{2}}{24}$

$\frac{M L^{2}}{12}$

$\frac{M L^{2}}{6}$

$\frac{\sqrt{2} M L^{2}}{24}$

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Explanation

To find the moment of inertia of a bent rod, we need to consider the moment of inertia of each half of the rod about the bending point and then add them. The moment of inertia of a rod about a perpendicular axis through its midpoint is (ML^2)/12.

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.