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NEET
Exam
NEET
Subject
Physics
Chapter
Motion in Two Dimensions

NEET · Physics · Motion in Two Dimensions

Two balls A and B are projected from the same position with...

Two balls AA and BB are projected from the same position with horizontal velocity 20ms−120m{s}^{- 1} and 5ms−15m{s}^{- 1} as shown. The separation between these two bodies at the moment their velocities become perpendicular to each other is xmxm. The value of xx is

Two balls A and B are projected from the same position with horizontal velocity 20m s ^ - 1 and 5m s ^ - 1 as shown.

Options

  1. A20
  2. B25
  3. C10
  4. D15

Correct answer

B. 25

Step-by-step solution

\overset{\to}{{v}_{1}} = 5\overset{^}{i} - gt\overset{^}{j} ⇒ {u}_{1} = 5

\overset{\to}{{v}_{2}} = 20\overset{^}{i} - gt\overset{^}{j} ⇒ {u}_{2} = 20

As v1→\overset{\to}{{v}_{1}} and v2→\overset{\to}{{v}_{2}} are perpendicular, 

v→1⋅v→2=0\overset{\to}{v}1 \cdot \overset{\to}{v}2 = 0
t=u1u2gt = \dfrac{\sqrt{{u}_{1}{u}_{2}}}{g}
s→rel=u→rel×t{\overset{\to}{s}}_{\text{rel}} = {\overset{\to}{u}}_{\text{rel}} \times t
=(u1+u2)t= ( {u}_{1} + {u}_{2} ) t
=(u1+u2)u1u2g= \left({u}_{1} + {u}_{2}\right)\dfrac{\sqrt{{u}_{1}{u}_{2}}}{g}
=25×1010=25m= 25 \times \dfrac{10}{10} = 25m

 
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