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(If the velocities are not along the same direction, (As the ball returns to its initial position, the change
→
(d) horizontal range of the projectile
→
Since s 1 = s 2 = d and s net = s + s | = 0
you'll need to use the vector from of this equations,
derived later in this section.) It's important to note the
in position, the change in position vector of the ball,
that is the net displacement will be zero).
order of the double subscripts in equation (1) v A/B
→
always means "velocity of A relative to B." These ∴ | VaV
| = 0.
subscripts obey an interesting kind of algebra, as
equation (1) shown. If regard each one as a fraction,
2. A long belt is moving horizontally with a speed of 4
then the fraction on the left side is the product of the
Km/hour. A child runs on this belt to and fro with a
fractions on the right sides : P/A = (P/B) (B/A). This
speed of 9 Km/hour (with respect to the belt) between
is a handy rule you can use when applying Equation
his father and mother located 50 m apart on the
(1) to any number of frames of reference. For
moving belt. For an observer on a stationary platform
example, if there are three different frames of
outside, what is the
reference A, B, and C, we can write immediately. (a) speed of the child running in the direction of motion
v P/A = v P/C + v C/B + V B/A
of the belt,
Step 4 : Evaluate your answer : Be on the lookout for
stray minus signs in your answer. If the target
(b) speed of the child running opposite to the direction of
motion of the belt and
variable is the velocity of a car relative to a bus
(v V/B ), make sure that you haven't accidentally
calculated the velocity of the bus relative of the car
(c) time taken by the child in case (a) and (b) ?
Which of the answers change, if motion is viewed by
one of the parents ?
(v B/C ). If you have made this mistake, you can
recover using equation.
v A/B = – v B/A
Sol. Let us consider positive direction of x-axis from left
to right
(a) Here, v B = + 4 Km/hour
Speed of child w.r.t. belt, v C = = 9 Km/hour
Solved Examples
∴ Speed of child w.r.t. stationary observer,
v C ′ = v C + v B or v C ′ = 9 + 4 = 13 Km/hour
1. A small glass ball is pushed with a speed V from A.
It moves on a smooth surface and collides with the
wall at B. If it loses half of its speed during the
collision, find the distance, average speed and
velocity of the ball till it reaches at its initial position.
(b) Here, v B = + 4 Km/hour, v C = – 9 Km/hour
∴ Speed of child w.r.t. stationary observer,
v C ′ = v C + v B or v C ′ = – 9 + 4 = –5 Km/hour
The negative sign shows that the child appears to run
in a direction opposite to the direction of motion of
the belt.
A V 0.5V B
(c) Distance between the parents, s = 50 m = 0.05 Km
Since parents and child are located on the same belt,
the speed of the child as observe by stationary
d
Sol. The ball moves from A to B with a constant speed V.
Since it loses half of its speed on collision, it returns
observer in either direction (either father to mother or
from mother to father) will be 9 Km/hour.
Time taken by the child in case (a) and (b),
from B to A with a constant speed V/2.
0.50 km
∴ V 1 = V and V 2 = V/2
t = = 20 sec.
9 km / hour
d1
+ d 2
Using the formula, V aV =
If the motion is observed by one of parents, answer to
(d1
/ V 1)
+ (d 2 / V2
)
case (a) case (b) gets altred. It is because the speed
Putting d 1 = d 2 = d; V 1 = V and V 2 = V/2
of the child w.r.t. either of mother or father is
d1
+ d 2 2V
9 Km/hour.
We obtain, V aV =
=
(dV) + (d / 0.5V) 3
3. A particle is projected with velocity v
From the formula,
0 = 100 m/s at
an angle θ = 30º with the horizontal. Find :
→ → →
→
| s1+
s2
| | snet
| (a) velocity of the particle after 2 sec.
average velocity = VaV
= =
s1
s2
t
(b) angle between initial velocity and the velocity after 2 sec.
+
net
V1
V2
(c) the maximum height reached by the projectile
| 1 2
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