Test: Work Power and Energy (August 9) - NEET MCQ

Gravitational potential energy of ball gets converted into elastic potential energy of the spring.

Net work done

Let the initial velocity of the shell be v, then by the conservation of momentum mv = Mv' where v' = velocity of gun.

Now, total K.E. 

But total K.E. = 1.05 kJ = 1.05 × 10³J

∴ v = 10² = 100 ms^–1.

When the body is thrown upwards. its K.E is converted  into P.E. The loss of energy due to air friction is the difference of K.E and P.E.

 = 200 – 180 = 20 J

The potential energy of a system increase if work is done by the system against a conservative force.
−ΔU = Wconservative force

As the two masses stick together after collision, hence it is inelastic collision.
Therefore, only momentum is conserved.

 Answer :- c

Solution :- Here initially sphere A is at rest and B is moving along x axis with velocity V.

After collision, velocity of sphere B becomes V/2 along direction perpendicular to its initial direction i.e. along Y axis.

Sphere A will move with some velocity u at some angle θ as shown in figure.considering law of conservation of momentum along X−axis.

m2V=m1ucosθ...........(i)

Considering law of conservation of momentum along Y−axis.

m2v/2=m1usinθ............(ii)

From equation i and ii.

1/2=tanθ

θ=tan−1(0.5).

So work done by the given force

 m₃v₃ = 20 (momentum of third part)

We have to use the Law of momentum conservation:

P ( initial ) = P ( final )

0 = n · m · u + ( M - n · m ) · v

where: n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg.

0 = 10 · 0.01 kg · 800 m/s + ( 100 kg - 10 · 0.01 kg ) · v

v = - 80 kgm/s / 99.9 kgm/s

v = 0.8 m/s

Then : F = Δ P / Δ t = ( 10 · 0.01 kg · 800 m/s ) : 5 s = 16 N

Answer: The average force exerted on the person is 16 N.

(t₁ = 1 minute; t₂ = 30 second given)