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TWo particles are located on a horizontal plane at a distance 60 m. At t=0, both the particles are simultaneously projected at angle 45 degree with velocities 2m/s and 14m/s respectively. Find (a) Minimum seperation between them during motion; (b) At what time is the seperation between them minimum?

TWo particles are located on a horizontal plane at a distance 60 m. At t=0, both the particles are simultaneously projected at angle 45 degree with velocities 2m/s and 14m/s respectively. Find (a) Minimum seperation between them during motion; (b) At what time is the seperation between them minimum?

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Grade:11

1 Answers

Arun
25750 Points
3 years ago
In relative motion, observer considers himself at rest and observes the motion of object. Graphically, we can draw the direction of motion of particle 2 w.r.t. particle l.
Both the particles are moving in gravitation field with same acceleration g. Hence, relative acceleration of particle 2 as seen from particle 1 will be zero. It means the relative velocity of the particle 2 w.r.t. particle 1 will be constant and will be equal to initial relative velocity. Graphically, we can draw the situation as shown in Fig.
AN is the minimum separation between the particles and BN is the relative separation between the particles when the distance between 1 and 2 is shortest. From figure, we can write
12
​ 
 cosθ=14cos45 
o
 +2cos45 
o
                     ...(i)
12
​ 
 sinθ=14 sin45 
o
 −2sin45 
o
                  ...(ii)
From (i) and (ii), v 
12
​ 
 =10 
2
​ 
 ms 
−1
 
cosθ= 
5
4
​ 
  and sinθ= 
5
3
​ 
 , as θ=37 
o
 
Hence, minimum separation between the particles
=AN=ABsinθ=60× 
5
3
​ 
 =36m
The time when separation between the particles is minimum,
t= 
∣ 
v
  
12
​ 
 ∣
BN
​ 
 ⇒t= 
10 
2
​ 
 
60cos37 
o
 
​ 
 = 
5
12 
2
​ 
 
​ 
 s

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