Question

A projectile is projected with speed $v$ at an angle $60^\circ$ with the horizontal. Find the ratio of the difference of kinetic energies at point $C$ (at ground) and point $B$ (at highest point) with the kinetic energy at point $C$ as shown in the diagram:

• A. $3:4$
• B. $1:3$
• C. $1:2$
• D. $1:12$

Hint:

At the highest point of a projectile, only horizontal velocity remains. Use $v\cos\theta$ directly to compute kinetic energy at the top.

Answer 1

The Correct Option is A

To solve this problem, we need to find the ratio of the difference in kinetic energies at two different points of a projectile's motion: point C (at the ground) and point B (at the highest point), in relation to the kinetic energy at point C.

First, let's understand the scenario:

• Point C is at the ground where the projectile is launched and lands again. At this point, the projectile has its maximum speed \(v\) because of the conversion of potential energy to kinetic energy.
• Point B is at the highest point of the projectile's trajectory. Here, the vertical component of the velocity is zero, and only the horizontal component contributes to the kinetic energy.

Let's calculate the kinetic energies at these points:

1. Kinetic Energy at Point C (K_C):
   The total initial kinetic energy when the projectile is launched can be calculated as: \(K_C = \frac{1}{2} m v^2\) where \(m\) is the mass of the projectile and \(v\) is the initial speed.
2. Kinetic Energy at Point B (K_B):
   At the highest point, only the horizontal component \(v_x\) contributes to the kinetic energy. The horizontal component is: \(v_x = v \cos 60^\circ = \frac{v}{2}\)
   So, the kinetic energy at B is: \(K_B = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{8} m v^2\)

Now, we find the difference in kinetic energy between points C and B:

• \(K_C - K_B = \frac{1}{2} m v^2 - \frac{1}{8} m v^2 = \frac{4}{8} m v^2 - \frac{1}{8} m v^2 = \frac{3}{8} m v^2\)

Finally, we calculate the ratio:

1. Ratio: 
   The ratio of the difference in kinetic energies to the kinetic energy at point C is: \(\frac{\frac{3}{8} m v^2}{\frac{1}{2} m v^2} = \frac{3}{8} \div \frac{4}{8} = \frac{3}{4}\) or \(3:4\)

Therefore, the ratio of the difference of kinetic energies at point C and point B with the kinetic energy at point C is 3:4.