Engineering math's ME 2

Engineering math's ME 2
Question:-
Option (A)
3 V cos &θ
Option (B)
2 V cos &θ
Option (C)
3/2 V cos &θ
Option (D)
Correct Option:
A
Question Solution:
At the highest point only horizontal component of velocity exists.
Let x be the peak point

Apply conservation of momentum principle we have
Momentum before explosion is equal to momentum after explosion
therefore, mV cos &θ = momentum before explosion
= momentum after explosion
Where
V1 = Velocity of second piece
After explosion half of mass traces back the path and hits the cannon, so its velocity will be equal to V cos &θ but in opposite direction.
therefore, From momentum conservation
mV cos &θ =
therefore, V1 = 3 V cos &θ
Question:-
Option (A)
at the top of circle
Option (B)
half way down from the top
Option (C)
quarter down from the top
Option (D)
at the bottom of circle
Correct Option:
D
Question:-
Option (A)
Zero
Option (B)
Option (C)
2 P
Option (D)
Correct Option:
B
Question Solution:
Resultant force in horizontal direction
= (3P - P)
= 2P towards left

Resultant force in vertical direction
= (4P - 2P) = 2P downwards
Overall resultant force
=
Question:-
Option (A)
forces are collinear
Option (B)
forces act at right angles to each other
Option (C)
forces are inclined at 60° to each other
Option (D)
forces can have any angle of inclination between them
Correct Option:
D
Question Solution:
R2= P2 + Q2 + 2PQ cos &θ
S2 = P2 + (-Q)2 + 2P(-Q)cos &θ
S2 = P2 + Q2 - 2PQ cos &θ
Adding we get,
R2 + S2 = 2(P2 + Q2)
Hence for a given identity of hold good, the forces can have any angle of inclination between them.