A ball of mass $200\, g$ rests on a vertical post of height $20\, m$ A bullet of mass $10 \,g$, travelling in horizontal direction, hits the centre of the ball After collision both travels independently The ball hits the ground at a distance $30 \,m$ and the bullet at a distance of $120 \,m$ from the foot of the postThe value of initial velocity of the bullet will be (if $g=10\, m / s ^2$ ) :
- $400\, m / s$
- $360 \,m / s$
- $60 \,m / s$
- $120 \,m / s$
The Correct Option is B
Solution and Explanation
The correct answer is (B) : 360 m/s
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Concepts Used:
Conservation of Energy
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
It also means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another. For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes.
So, mathematically we can represent the law of energy conservation as the following,
The amount of energy spent in a work = The amount of Energy gained in the related work
Now, the derivation of the energy conservation formula is as followed,
Ein − Eout = Δ Esys
We know that the net amount of energy which is transferred in or out of any system is mainly seen in the forms of heat (Q), mass (m) or work (W). Hence, on re-arranging the above equation, we get,
Ein − Eout = Q − W
Now, on dividing all the terms into both the sides of the equation by the mass of the system, the equation represents the law of conservation of energy on a unit mass basis, such as
Q − W = Δ u
Thus, the conservation of energy formula can be written as follows,
Q – W = dU / dt
Here,
Esys = Energy of the system as a whole
Ein = Incoming energy
Eout = Outgoing energy
E = Energy
Q = Heat
M = Mass
W = Work
T = Time