A bullet is fired at a target with a velocity of 80 m/s and penetrates 50 cm into it. If this bullet were fired into…

A bullet is fired at a target with a velocity of 80 m/s and penetrates 50 cm into it. If this bullet were fired into a target 25 cm thick with equal velocity, with what velocity would it emerge, supposing the resistance is uniform and same in both the cases?

(a) √80 m/s

(b) 40/√2 m/s

(c) 40 m/s

(d) 40√2 m/s

Ans: (d)

Sol:

Applying , \displaystyle v^2 = u^2 + 2 a S

\displaystyle 0 = 80^2 + 2 a (0.5)

a = – 6400 m⁄s2

Again , \displaystyle v^2 = u^2 + 2 a S

\displaystyle v^2 = 80^2 + 2 (-6400) (0.25)

\displaystyle v^2 = 6400  - 3200 = 3200

\displaystyle v = \sqrt{3200} = 40 \sqrt{2} m/s

A ball is allowed to fall down with initial speed v from a height of 10 m. It loses 50% kinetic energy…

Q: A ball is allowed to fall down with initial speed v from a height of 10 m. It loses 50% kinetic energy after striking the floor. It reaches to the same height after collision. What is the value of v ?

(a) 28 m/s

(b) 7 m/s

(c) 14 m/s

(d) it is never possible

Ans: (c)

Sol:

To reach same height of 10 m after collision,

Velocity After collision \displaystyle = \sqrt{2 g h}

\displaystyle = \sqrt{2 \times 10 \times 10} = 10\sqrt{2} m/s

As energy lost in collision is 50% , velocity becomes v⁄√2

∴ Velocity of ball on striking,

v = (10 √2) √2 = 20 m/s

Applying \displaystyle v^2 = u^2 + 2 a S

\displaystyle 20^2 = u^2 + 2 \times 10 \times 10

\displaystyle u^2 = 20^2 - 200 = 200

\displaystyle u = \sqrt{200} = 10\sqrt{2}

u = 14.14 m/s

A ball of mass ‘m’ moving with a horizontal velocity ‘v’ strikes the bob of mass ‘m’ of a pendulum at rest. ..

Q: A ball of mass ‘m’ moving with a horizontal velocity ‘v’ strikes the bob of mass ‘m’ of a pendulum at rest. During this collision, the ball sticks with the bob of the pendulum. The height to which the combined mass raises is (g = acceleration due to gravity)

(a) \displaystyle \frac{v^2}{4 g}

(b) \displaystyle \frac{v^2}{8 g}

(c) \displaystyle \frac{v^2}{ g}

(d) \displaystyle \frac{v^2}{2 g}

Ans: (b)

Sol: Applying conservation of momentum ,

m v = (m + m)v’

v’ = v/2

Applying conservation of Energy ,

\displaystyle  \frac{1}{2} (2 m) v'^2 = (2 m) g h

\displaystyle  h = \frac{v'^2}{2 g}

\displaystyle h = (\frac{v}{2})^2 \times \frac{1}{2 g}

\displaystyle  h = \frac{v^2}{8 g}