Two objects, x and y, are held at rest on a horizontal frictionless surface and a spring is compressed between them. the mass of x is 2/5 times the mass of y. immediately after the spring is released, x has a kinetic energy of 50 j and y has a kinetic energy of:

20J

Explanation:

Using conservation law of momentum;

since the bodies were at rest, their initial momentum is zero

0 =  M1Vx + M2Vy

- M1Vx = M2Vy where Vx is the final velocity of x after the spring has been release and Vy is final velocity of y and M1 and M2 are the masses of x and y

also M1 = 2/5 M2

substitute M1 into the the equation above

-2/5 M2Vx =  M2Vy

cancel M2 on both side

-2/5Vx =  Vy

comparing the kinetic energy of both x and y

for x K.E = 1/2 M1 Vx²

and y K.E = 1/2M2 Vy²

substitute for M1 = 2/5 M2

K.Ex = 1/2 × 2/5 M2 Vx²

divide  K.Ex / K.Ey = (1/2 × 2/5 M2 Vx²) / 1/2 M2 Vy²

cancel the common terms

K.Ex / K.Ey = (2/5 Vx²) / Vy²

substitute -2/5Vx for Vy

(2/5 Vx²) / ( -2/5 Vx)² = (2/5 Vx²) / ( 4/25 Vx²)

cancel Vx²

(2/5) / (4/25) = 2/5 ÷ 4/25 = 2/5 × 25/4 = 5/2

the ratio of x and y kinetic energy is 5:2

since the kinetic energy of x is 50

50 : 20 = 5 : 2 if 10 is used to divide both sides

the kinetic energy of y = 20 J

no sign of dia of fuel cell