Two particles move in a uniform gravitational field with an acceleration g.At the initial moment the particles were located at same point and moved with velocities u_(1)=9 ms^(-1) and u_(2)=4 ms^(-1) horizontally in opposite directions.The time between the particles at the moment when their velocity vectors are mutually perpendicular in s in (take g=10 ms^(-2))
Answers
Explanation:
1.11 Two particles move in a uniform gravitational field with an accelaration g. At the initial
moment the particles were located at one point and moved with velocities v1 = 3.0ms−1 and
v2 = 4.0ms−1 horizontally in opposite directions. Find the distance between the particles at the
moment when their velocities become mutually perpendicular.
Sol. We can visualize the situation as given below. Suppose that the velocities of two bodies
become mutually perpendicular after time t. Let the velocities of the two bodies at that instant be
v
0
1
and v
0
2
, respectively. What can we say about these velocities?
v
0
2
v
0
1
v2 v1
v2t v1t +i
+j
Thought Process
• The velocity of each body can be resolved into horizontal and vertical component (ˆi and ˆj).
• The accelaration due to gravity g acts only on the vertical component of velocity. The
horizontal component remains unchanged for both bodies.
• Since the initial velocity in the vertical direction is 0 for both bodies, they travel the same
distance in vertical direction. Hence, the distance between the bodies at t is given only by
the horizontal distance between the bodies at time t.
We have,
v
0
1 = v1
ˆi − gtˆj
v
0
2 = −v2
ˆi − gtˆj
At time t, the vectors v
0
1
and v
0
2
are mutally perpendicular. Hence, their dot product must be
0.
v
0
1
· v
0
2 =0
(v1
ˆi − gtˆj) · (−v2
ˆi − gtˆj) =0
−v1v2
ˆi · ˆi − v1gtˆi · ˆj − v2gtˆi · ˆj + (gt)
2ˆj · ˆj =0
−v1v2 + (gt)
2 =0 [∵ ˆi · ˆi = 1 and ˆi · ˆj = 0]
t =
√
v1v2
g
In time t, the bodies have moved v1t and v2t in opposite directions, therefore the distance
between them is
(v1 + v2)t = (v1 + v2)
√
v1v2
g
