A ball of mass 200 ~g rests on a vertical post of height 20 ~m . A bullet of mass 10 ~g , travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 ~m and the bullet at a distance of 120 ~m from the foot of the post. The value of initial velocity of the bullet will be (if g=10 ~m / s^2 ) :

Center of Mass PYQ — JEE Main 2023 · 30 Jan · Shift 1 · Q56 with Solution

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Correct answer: B

Given data

Mass of ball: $M = 200\text{ g} = 0.2\text{ kg}$
Mass of bullet: $m = 10\text{ g} = 0.01\text{ kg}$
Height of post: $h = 20\text{ m}$
Horizontal distance travelled by ball after collision: $x_1 = 30\text{ m}$
Horizontal distance travelled by bullet after collision: $x_2 = 120\text{ m}$
$g = 10\text{ m/s}^2$

The bullet hits the centre of the ball horizontally, so just after collision both the ball and bullet have horizontal velocities only.

Time taken to fall from height $20\text{ m}$

For both ball and bullet, vertical motion starts with zero vertical velocity from the same height.

h = \frac{1}{2}gt^2

20 = \frac{1}{2}(10)t^2 = 5t^2

t^2 = 4 \quad \Rightarrow \quad t = 2\text{ s}

So, each reaches the ground after $2\text{ s}$ .

Horizontal velocities after collision

For the ball:

v_1 = \frac{x_1}{t} = \frac{30}{2} = 15\text{ m/s}

For the bullet:

v_2 = \frac{x_2}{t} = \frac{120}{2} = 60\text{ m/s}

So after collision,

ball moves with horizontal speed $15\text{ m/s}$ ,
bullet moves with horizontal speed $60\text{ m/s}$ .

Apply conservation of horizontal momentum

Let initial velocity of bullet be $u$ .

Initially only bullet is moving, so initial horizontal momentum is:

mu

After collision, horizontal momentum is:

Mv_1 + mv_2

Thus,

mu = Mv_1 + mv_2

Substitute values:

0.01u = 0.2(15) + 0.01(60)

0.01u = 3 + 0.6 = 3.6

u = \frac{3.6}{0.01} = 360\text{ m/s}

Final answer

\boxed{360\text{ m/s}}

So the correct option is B.

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View written solutionFree

Correct answer: B

Given data

Mass of ball: $M = 200\text{ g} = 0.2\text{ kg}$
Mass of bullet: $m = 10\text{ g} = 0.01\text{ kg}$
Height of post: $h = 20\text{ m}$
Horizontal distance travelled by ball after collision: $x_1 = 30\text{ m}$
Horizontal distance travelled by bullet after collision: $x_2 = 120\text{ m}$
$g = 10\text{ m/s}^2$

The bullet hits the centre of the ball horizontally, so just after collision both the ball and bullet have horizontal velocities only.

Time taken to fall from height $20\text{ m}$

For both ball and bullet, vertical motion starts with zero vertical velocity from the same height.

h = \frac{1}{2}gt^2

20 = \frac{1}{2}(10)t^2 = 5t^2

t^2 = 4 \quad \Rightarrow \quad t = 2\text{ s}

So, each reaches the ground after $2\text{ s}$ .

Horizontal velocities after collision

For the ball:

v_1 = \frac{x_1}{t} = \frac{30}{2} = 15\text{ m/s}

For the bullet:

v_2 = \frac{x_2}{t} = \frac{120}{2} = 60\text{ m/s}

So after collision,

ball moves with horizontal speed $15\text{ m/s}$ ,
bullet moves with horizontal speed $60\text{ m/s}$ .

Apply conservation of horizontal momentum

Let initial velocity of bullet be $u$ .

Initially only bullet is moving, so initial horizontal momentum is:

mu

After collision, horizontal momentum is:

Mv_1 + mv_2

Thus,

mu = Mv_1 + mv_2

Substitute values:

0.01u = 0.2(15) + 0.01(60)

0.01u = 3 + 0.6 = 3.6

u = \frac{3.6}{0.01} = 360\text{ m/s}

Final answer

\boxed{360\text{ m/s}}

So the correct option is B.

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