Question

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 postThe value of initial velocity of the bullet will be (if $g=10\, m / s ^2$ ) :

• A. $400\, m / s$
• B. $360 \,m / s$
• C. $60 \,m / s$
• D. $120 \,m / s$

Answer 1

The Correct Option is B

To solve this problem, we need to understand the dynamics of the collision and the subsequent motion of both the ball and the bullet. Both objects travel under the influence of gravity after the collision.

Initial conditions before the collision:

• \(m_b = 200 \, \text{g} = 0.2 \, \text{kg}\) (mass of the ball)
• \(m_bu = 10 \, \text{g} = 0.01 \, \text{kg}\) (mass of the bullet)
• \(h = 20 \, \text{m}\) (height of the post)
• The initial velocity of the bullet is \(u_b\) which we need to determine.

After the collision, the horizontal distance traveled by the ball and bullet are given:

• Ball: \(d_b = 30 \, \text{m}\)
• Bullet: \(d_bu = 120 \, \text{m}\)

Since the collision is assumed to be perfectly elastic or inelastic in method and both bodies move horizontally off the top of the post, they will have the same time of flight, which is solely determined by the height of fall \(h\).

The time \(t\) to fall a height \(h\) can be calculated using the equation:

\(h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2h}{g}}\)

\(t = \sqrt{\frac{2 \times 20}{10}} = \sqrt{4} = 2 \, \text{s}\)

The horizontal velocity of each object determines how far they travel horizontally. Using the formula for horizontal range \(R = v_x \cdot t\):

For the ball:

\(v_b = \frac{d_b}{t} = \frac{30}{2} = 15 \, \text{m/s}\)

For the bullet:

\(v_bu = \frac{d_bu}{t} = \frac{120}{2} = 60 \, \text{m/s}\)

Assuming the collision conserves linear momentum, we write the conservation equation for the system:

\(m_bu \cdot u_b + m_b \cdot 0 = m_b \cdot v_b + m_bu \cdot v_bu\)

Substitute the known masses and velocities:

\(0.01 \cdot u_b = 0.2 \cdot 15 + 0.01 \cdot 60\)

Simplify and solve for \(u_b\):

\(0.01 \cdot u_b = 3 + 0.6 = 3.6\)

\(u_b = \frac{3.6}{0.01} = 360 \, \text{m/s}\)

Therefore, the initial velocity of the bullet should be \(360 \, \text{m/s}\).