Question

A 6.0-kg object, initially at rest in free space, "explodes" into three segments of equal mass. Two of these segments are observed to be moving with equal speeds of 20 m/s with an angle of 60° between their directions of motion. How much kinetic energy is released in this explosion?

Answer 1

Answer:

$Q = 2000 J$

Explanation:

As we know that the 6 kg object was at rest initially

So here since net force on the system is zero

so the momentum of the system will always remains conserved

so we can say

$0 = P_1 + P_2 + P_3$

now we know that

$P_1 = P_2 = P$

and the angle between the two objects is 60 degree

so we can say

$\vec P_1 + \vec P_2 = \sqrt{P_1^2 + P_2^2 + 2P_1 P_2cos60}$

$\vec P_1 + \vec P_2 = \sqrt{P^2 + P^2 + 2P^2(0.5)} = \sqrt3 P$

now we can say that the speed of the third mass will be

$v_3 = \sqrt 3 (20) m/s$

now the total kinetic energy released in this system is given as

$Q = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 + \frac{1}{2}mv_3^2$

$Q = \frac{1}{2}(2)(20^2) + \frac{1}{2}(2)(20^2) + \frac{1}{2}(2)(20\sqrt3)^2$

$Q = 2000 J$