The one which does not represent a force in any context is
Friction
Impulse
Tension
Weight
B.
Impulse
Impulse is equal to the change in momentum. So, option (b) is the correct answer.
Water from a hose pipe of radius 5 cm strikes a wall normally at a speed of 5 ms-1. The force exerted on the wall in newton is
C.
Given, Radius of hose pipe (r) = 5 cm
= 5 × 10-2 m
Speed of water (v) = 5 ms-1
We know that,
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc 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 = mv1 + 2mv2 .....(i)
v2 -v1 = v ...... (ii)
Therefore, Collision is elastic, e = 1
From equ (i) and equ (ii) v1 = -v/3
Now, for the collision of neutron with carbon nucleus
Applying conservation of momentum
mv + 0 = mv1 + 12mv2 ....; (iii)
v = v2-v1 ....(iv)
A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 m/s2 )
2.0
4.0
1.6
1.6
A.
2.0
Let the mass of block be m.
Frictional force in rest position
F = mg sin 30
Two masses m1 = 5 kg and m2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift free to move?
(g = 9.8 m/s2 )
0.2 m/s2
9.8 m/s2
5 m/s2
5 m/s2
A.
0.2 m/s2
On release, the motion of the system will be according to the figure.
m1g - T = m1a ...... (i) and
T - m2g = m2a ...... (ii)
On solving;
Here, m1 = 5 kg, m2 = 4.8 kg, g = 9.8 m/s2
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