A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of √2gh. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of √(h/g) is :

Question

A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of √2gh. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of √(h/g) is : (A) √(1/2) (B) (1/2) (C) √(3/2) (D) √(3/4)

Tardigrade Admin · Accepted Answer

Particles will collide after time t0=(h/√2 gh) at collision, vA=gt0 vB=uB-gt0 ⇒ vA=-vB Time taken by combined mass to reach theground time =√(2×3h / 4/g)=√(3h/2g)

$\frac{1}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{3}{4}$

Particles will collide after time $t_{0} = \frac{h}{2 g h}$
at collision, $v_{A} = g t_{0} v_{B} = u_{B} - g t_{0}$
$\Rightarrow v_{A} = - v_{B}$
Time taken by combined mass to reach theground
time $= \frac{2 \times 3 h /4}{g} = \frac{3 h}{2 g}$

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Particles will collide after time t0​=2gh​h​ at collision, vA​=gt0​vB​=uB​−gt0​ ⇒vA​=−vB​ Time taken by combined mass to reach theground time =g2×3h/4​​=2g3h​​

$\frac{1}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{3}{4}$

Particles will collide after time $t_{0} = \frac{h}{2 g h}$
at collision, $v_{A} = g t_{0} v_{B} = u_{B} - g t_{0}$
$\Rightarrow v_{A} = - v_{B}$
Time taken by combined mass to reach theground
time $= \frac{2 \times 3 h /4}{g} = \frac{3 h}{2 g}$