An object is projected with kinetic energy $K$ from a point $A$ at an angle $60^{\circ}$ with the horizontal. The ratio of the difference in kinetic energies at points $B$ and $C$ to that at point $A$ (see figure), in the absence of air friction is :

Question

An object is projected with kinetic energy

K

from a point

A

at an angle

60^{\circ}

with the horizontal. The ratio of the difference in kinetic energies at points

B

and

C

to that at point

A

(see figure), in the absence of air friction is :
An object is projected with kinetic energy K from a point A at an angle 60^ with the horizontal. The ratio of the

An object is projected with kinetic energy K from a point A at an angle 60^ with the horizontal. The ratio of the

Quantrex Academy · Accepted Answer

Correct answer: C. 3: 4

At A: KE_A = K , v_A = v , angle = 60° . At B (highest point): v_B = v 60° = v/2 , so KE_B = K/4 . At C (same level as A): by energy conservation, KE_C = K . |KE_B - KE_C| = |K/4 - K| = 3K/4 . Ratio = (3K/4)/(K) = (3)/(4) , i.e., 3:4 .

An object is projected with kinetic energy K from a point A at an...

Options

Correct answer

Step-by-step solution

More from Motion in Two Dimensions