Question 147: Centre of Mass and Rotational Motion

Question

A particle of mass m is projected with a speed
u from the ground at an angle
$$	heta  = {\pi  \over 3}$$ w.r.t.
horizontal (x-axis). When it has reached its
maximum height, it collides completely
inelastically with another particle of the same
mass and velocity $$u\widehat i$$ . The horizontal distance
covered by the combined mass before reaching
the ground is:

Accepted Answer

By momentum conservation,
$${{mu} \over 2}$$ + mu = 2mV
$$ \Rightarrow $$ V = $${{3u} \over 4}$$
H
max
= $${{{u^2}{{\sin }^2}60^\circ } \over {2g}}$$ = $${{{u^2} 	imes {3 \over 4}} \over {2g}}$$ = $${{3{u^2}} \over {8g}}$$
Time taken = $$\sqrt {{{2{H_{\max }}} \over g}} $$ = $$\sqrt {{2 \over g} 	imes {{3{u^2}} \over {8g}}} $$ = $${{\sqrt 3 } \over 2}{u \over g}$$
Horizontal distance traveled = ut
= $${{3u} \over 4}.{{\sqrt 3 } \over 2}{u \over g}$$ = $${{3\sqrt 3 } \over 8}{{{u^2}} \over g}$$

Answer

$$2\sqrt 2 {{{u^2}} \over g}$$

Answer

$${{3\sqrt 3 } \over 8}{{{u^2}} \over g}$$

Answer

$${{3\sqrt 2 } \over 4}{{{u^2}} \over g}$$

Answer

$${5 \over 8}{{{u^2}} \over g}$$

Centre of Mass and Rotational Motion

Choose the correct answer: