Question

An object with mass 3M is launched straight up. When it reaches its maximum height, a small explosion breaks the object into two pieces with masses M and 2M. The explosion provides an impulse to each object in the horizontal direction only. The heavier of the two pieces is observed to land a distance 53 m from the point of the original launch. How far from the original launch position does the lighter piece land

Answer 1

Answer:

$x_{L} = 106\,m$

Explanation:

Each piece is analyzed by using the Principle of Momentum Conservation and the Impulse Theorem:

Heavier object:

$(2\cdot M)\cdot (0) + F\cdot \Delta t = (2\cdot M)\cdot v_{H}$

Lighter object:

$M\cdot (0) + F\cdot \Delta t = M\cdot v_{L}$

After the some algebraic handling, the following relationship is found:

$M\cdot v_{L} = 2\cdot M\cdot v_{H}$

$v_{L} = 2\cdot v_{H}$

Given that both pieces have horizontal velocities only and both are modelled as projectiles, the horizontal component of velocity remains constant and directly proportional to travelled distance. Then:

$\frac{v_{L}}{v_{H}} = \frac{x_{L}}{53\,m}$

$2 = \frac{x_{L}}{53\,m}$

$x_{L} = 106\,m$