Question

A projectile is launched vertically upward. It explodes into two pieces at the top point of its trajectory. One piece has twice the mass of the other. Immediately after the explosion the more massive piece has kinetic energy $E$. What is the total kinetic energy of both pieces immediately after the explosion?

• A. 1.5 E
• B. 2 E
• C. 3 E
• D. Not enough information to answer

Answer 1

Answer: (C)

3 E

There is no external force acting on the projectile in the horizontal direction.
Hence, momentum is conserved in the horizontal direction.
Let the masses be $m_1$ and $m_2$ respectively.
Let the velocities be $v_1$ and $v_2$ respectively.
$m_2=2m_1$
From conservation of momentum, $m_1{v}_1=2m_1{v}_2$
$\Rightarrow v_1=2v_2$
Given,

$\frac{1}{2}2m_1{v}_2^2=E$

Ke of the lighter fragment is $\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}\frac{m_2}{2}(2v_2{)}^2=m_2{v}_2^2=2(\frac{1}{2}{m}_2{v}_2^2)=2E$

Total KE: $2E+E=3E$