A machine gun fires a bullet of mass 40 g with a velocity 1200 ms−1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?
one
Four
Two
Two
D.
Two
The force exerted by machine gun on man's hand in firing a bullet = change in momentum per second on a bullet or rate of change of momentum
The force exerted by man on machine gun = 144 N Hence, number of bullets fired =144/48 = 3
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
600 m
B.
1000 m
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
60 m
D.
60 m
Third equation of motion gives
v2 = u2 + 2as ⇒ 2
s ∝ u (since v = 0)
where a = retardation of body in both the cases
therefore, .... (i)
Here, s1 = 20 m, u1 = 60 km/h, u2 = 120 km/h. Putting the given values in eq. (i), we get
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
8h/9 metres from the ground
C.
8h/9 metres from the ground
A mass ‘m’ moves with a velocity v and collides inelastically with another identical mass. After collision, the 1st 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
2v/√3
C.
2v/√3
mv= mv1 cos = θ