A ball of mass 0.2 kg rests on a vertical post of height 5 cm. A bullet of mass 0.01 kg, travelling with a velocity…

Q: A ball of mass 0.2 kg rests on a vertical post of height 5 cm. A bullet of mass 0.01 kg, travelling with a velocity v m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity v of the bullet is

Numerical

(a)250 m/s

(b)250√2 m/s

(c)400 m/s

(d)500 m/s

Ans: (d)

Sol: Time taken by ball & bullet to hit the ground is

$\large t = \sqrt{\frac{2h}{g}}$

$\large t = \sqrt{\frac{2\times 5}{10}}$

t = 1 sec

v1 × t = 20 m/s

v2 × t = 100

v2 = 100 m/s

Applying momentum conservation before & after collision

0.01 × v = 0.2 × 20 + 0.01 × 100

v = 500 m/s

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit…

Q: Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively, as shown in the figure between collisions, the particle move with constant speed. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?

Numerical

(a)4

(b)3

(c)2

(d)1

Ans: (c)

Sol: After head-on elastic collision , velocities will be exchanged . Since one particle speed has double the other , hence one particle will travel a distance S other will travel 2S . Therefore after two collisions , particle again collide at A .

Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness. The mass of ink used to draw…

Q: Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of two inner circles and each of the two line segments is m, the mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0,0), left inner circle (-a,a), right inner circle (a,a), vertical line (0,0) and horizontal line (0,-a). The y-coordinate of the centre of mass of the ink in this drawing is

Numerical

(a)a/10

(b)a/8

(c)a/12

(d)a/3

Ans: (a)

Sol: $\large Y_{com} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4 + m_5 y_5}{m_1 + m_2 + m_3 + m_4 + m_5}$

$\large Y_{com} = \frac{6m \times 0 + m \times a + m \times a + m \times 0 + m \times (-a)}{6m + m + m + m + m}$

$\large Y_{com} = \frac{a}{10} $

A pulse of light of duration 100 ns is absorbed completely a small object initially at rest…

Q: A pulse of light of duration 100 ns is absorbed completely a small object initially at rest. Power of the pulse is 30 mV and the speed of light is 3×108 ms-1. The final momentum of the object is

(a)0.3×10-17 kg-ms-1

(b)1.0×1017 kg-ms-1

(c)3.0×1017 kg-ms-1

(d)9.0×1017 kg-ms-1

Ans: (b)

Sol: Final Momentum of the Object $\large = \frac{Power \times time}{Speed \; of \; light}$

$\large = \frac{30 \times 10^{-3}\times 100 \times 10^{-9}}{3 \times 10^{8}}$

= 1.0 × 10-17 kg m/s

A particle of mass m is projected from the ground with an initial speed u0 at an angle a with the horizontal…

Q: A particle of mass m is projected from the ground with an initial speed u0 at an angle α with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u0 . The angle that the composite system makes with the horizontal after the collision is

(a) π/4

(b) $\frac{\pi}{4} + \alpha$

(c) $\frac{\pi}{4} -\alpha$

(d) π/2

Ans: (d)

Sol: Maximum height attained by the particle is

$\large H = \frac{u_0^2 sin^2 \alpha}{2 g}$ …(i)

Velocity of the particle at height H which was thrown vertically

$\large v = \sqrt{u_0^2 – 2 g H}$

Now , applying conservation of linear momentum

$\large m u_o cos\alpha \hat{i} + m(\sqrt{u_0^2 – 2 g H})\hat{j} = 2 m \vec{v’} $

$\large m u_o cos\alpha \hat{i} + m(\sqrt{u_0^2 – 2 g (\frac{u_0^2 sin^2 \alpha}{2 g})})\hat{j} = 2 m \vec{v’} $

$\large m u_o cos\alpha \hat{i} + m u_o cos\alpha \hat{j}= 2 m \vec{v’}$

$\large \vec{v’} = \frac{u_0 cos\alpha}{2}\hat{i} + \frac{u_0 cos\alpha}{2}\hat{j}$

$\large tan\theta = \frac{u_0 cos\alpha /2}{u_0 cos\alpha /2}$

tanθ = 1  ⇒ θ = 45°