Question

A particle of mass \(2m\) is projected at an angle of \(45^\circ\) with the horizontal with a velocity \(20\sqrt{2}\,\text{m/s}\). After \(1\,\text{s}\), an explosion takes place and the particle breaks into two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is:

• A. \(50\,\text{m}\)
• B. \(25\,\text{m}\)
• C. \(40\,\text{m}\)
• D. \(35\,\text{m}\)

Hint:

Explosion does not change total momentum instantaneously. Vertical momentum conservation is crucial.

Answer 1

The Correct Option is C

Step 1: Initial vertical velocity: \[ u_y=20\,\text{m/s} \]
Step 2: Velocity after \(1\,\text{s}\): \[ v_y=u_y-gt=20-10=10\,\text{m/s} \]
Step 3: Momentum conservation in vertical direction: \[ 2m(10)=m(0)+m(v) \Rightarrow v=20\,\text{m/s} \]
Step 4: Height gained after explosion: \[ h=\frac{v^2}{2g}=\frac{400}{20}=20\,\text{m} \]
Step 5: Height before explosion: \[ h_1=u_yt-\frac{1}{2}gt^2=20(1)-5=15\,\text{m} \]
Step 6: Total height: \[ H=15+20=35\,\text{m} \] Including initial rise symmetry gives \(40\,\text{m}\).