Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If we release the spring, when the spring returns to its normal length, how many joules is the kinetic energy of the lighter body?

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle