Question

An angry bird (flightless) of mass 5 kg is projected with a speed of 25 m/s at an angle of ${37}^{\circ }$ with the horizontal. When it reaches its maximum height, it screams "fire in the hole!” and decides to take a massive 1 kg dump on Ram standing directly below it. What will be the magnitude of the horizontal velocity of the bird after release?

• A. 20 m/s
• B. 10 m/s
• C. 25 m/s
• D. None of these

Answer 1

The correct option is C

25 m/s

At the highest point, we can conserve horizontal momentum as net force in the horizontal direction is zero. Therefore change in momentum in the horizontal direction will be zero.

Velocity of bird initially towards horizontal direction

= 25 cos ${37}^{\circ }$ = 25 $\times$ $\frac{4}{5}$ = 20 m/s

$\therefore P_i=m\times u=5\times 20$

= 100 kg m/s

Final horizontal momentum = $m_1{v}_1+m_2{v}_2$

= 1(0) + 4$\left(v_2\right)$

= $4v_{2}kgm/s$

As m, is moving vertically downward.

Its horizontal momentum will be zero and $v_2$ is the horizontal component of the final velocity of $m_2$.

$\therefore P_i=P_f$

$\Rightarrow 100=4v_2$

$\Rightarrow v_2=\frac{100}{4}=25im/s$

$\therefore$ Its horizontal speed will be 25 m/s

note that we cannot conserve momentum along the vertical direction as there is a net force in that direction i.e. gravity.