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
Three
D.
Three
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
2.0
4.0
1.6
2.4
A.
2.0
Let the mass of block be m.
Frictional force in rest position
F = mg sin 30
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is p_{d}; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p_{c}. The values of p_{d} and p_{c} are respectively :
(0,1)
(.89,.28)
(.28,.89)
(0,0)
B.
(.89,.28)
For collision of a neutron with deuterium:
Applying conservation of momentum:
mv + 0 = mv_{1} + 2mv_{2} .....(i)
v_{2} -v_{1} = v ...... (ii)
Therefore, Collision is elastic, e = 1
From equ (i) and equ (ii) v_{1} = -v/3
${P}_{d}=\frac{{\displaystyle \frac{1}{2}}m{v}^{2}-{\displaystyle \frac{1}{2}}m{v}_{1}^{2}}{{\displaystyle \frac{1}{2}}m{v}^{2}}=\frac{8}{9}=0.89$
Now, for the collision of neutron with carbon nucleus
Applying conservation of momentum
mv + 0 = mv_{1} + 12mv_{2} ....; (iii)
v = v_{2}-v_{1} ....(iv)
${v}_{1}=-\frac{11}{13}v\phantom{\rule{0ex}{0ex}}{P}_{c}=\frac{{\displaystyle \frac{1}{2}}m{v}^{2}-{\displaystyle \frac{1}{2}}m{\left({\displaystyle \frac{11}{13}}v\right)}^{2}}{{\displaystyle \frac{1}{2}}m{v}^{2}}=\frac{48}{169}\approx 0.28$
A car is negotiating a curved road of radius R. The road is banked at angle . The coefficient of friction between the tyres of the car and the road is . The maximum safe velocity on this road is,
A.
A car is negotiating a curved road of radius R. The road is banked at angle and the coefficient of friction between the tyres of car and the road is .
The given situation is illustrated as:
In the case of vertical equilibrium,
N cos = mg + f_{1} sin
mg = N cos ... (i)
In the case of horizontal equilibrium,
... (ii)
Dividing Eqns. (i) and (ii), we get