A mass ‘m’ moves with a velocity v and collides inelastically with another identical mass. After collision, the 1^{st} mass moves with velocity v/√3 in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision
v
√3 v
2v/√3
v/√3
C.
2v/√3
mv= mv_{1} cos = θ
Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is [µ =k 0.5]
800 m
1000 m
600 m
900 m
B.
1000 m
A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?
h/9 metres from the ground
7h/9 metres from the ground
8h/9 metres from the ground
17h/18 metres from the ground.
C.
8h/9 metres from the ground
An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be
20 m
40 m
60 m
80 m
D.
80 m
Third equation of motion gives
v^{2} = u^{2} + 2as ⇒ 2
s ∝ u (since v = 0)
where a = retardation of body in both the cases
therefore, .... (i)
Here, s_{1} = 20 m, u1 = 60 km/h, u_{2} = 120 km/h. Putting the given values in eq. (i), we get