Question

A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, traveling with a velocity V meter per second in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is:
A) 250 meter per second
B) 251.41 meter per second
C) 400 meter per second
D) 500 meter per second

Answer 1

In this question, we will use the third equation of motion which gives relation between the displacement, time, initial velocity and acceleration on an object. Further we will use the law of conservation of momentum to find the final velocity of the given bullet. Also, we will see the basics of the equation of motion and the law of conservation of momentum.
Formula used:
$s = ut + \dfrac{1}{2}g{t^2}$
$M \times {V_f} + m \times {v_f} = M \times {V_i} + m \times {v_i}$

Complete answer:
Let the mass of bullet m and the mass of ball be M. It is given that initially the ball is at a height of 5m at rest, so here; the only acceleration will be acceleration due to gravity. So here we apply the third equation of motion:
$s = ut + \dfrac{1}{2}g{t^2}$
By substituting the given values in the above equation:
\eqalign{ & 5 = 0 + \dfrac{1}{2}(10){t^2} \cr & \therefore t = 1\sec \cr}
Since, we have the velocity of ball v and velocity of bullet V. Further, by collision law of conservation of momentum comes into action, which is given by:
\eqalign{ & M \times {V_f} + m \times {v_f} = M \times {V_i} + m \times {v_i} \cr & \Rightarrow 0.01 \times {V_f} = 0.01 \times 100 + 0.20 \times 20 \cr & \therefore {V_f} = 500m/s.........({V_f} = V) \cr}

Therefore, option D) is the correct answer, i.e., the time t is 1 sec and the velocity of bullet is given by V in the above result.

Additional information:
As we know that the equations of motion are equations which describe the behavior of a physical system in terms of its motion as a function of time. Further we can say that these equations of motion describe the behavior of a physical system as a set of mathematical functions in terms of dynamic variables. Here, dynamic variables are said to be normally spatial coordinates and time is used, but others are also possible, like momentum components and time.
From the law of conservation of mass which says that mass can neither be created nor destroyed in any chemical reactions. Also, it can be defined as the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction.
We know that collision means when two objects come in contact with each other for a very short period. Collision is an interaction between two masses for a very short interval where the momentum and energy of the colliding masses changes. Here are two types of collision: first id elastic collision where the energy remains same after interaction or collision and second is inelastic collision where the final energy changes after the collision of the particles or body.

Note:
Here, we should remember that the three different equations of motion are used in finding different physical properties of a particle under motion. We should also observe that these equations are only applicable to the classical system not in the quantum system. Further, during the solution, one should see that laws of conservation are not violated, i.e., mass can neither be created nor be destroyed and similarly for energy of the particles as well.