A1.0 kg lump of clay is sliding to the right on a frictionless surface with speed 2 m/s . it collides head-on and sticks to a 0.5 kg metal sphere that is sliding to the left with speed 4 m/s . what is the kinetic energy of the combined objects after the collision?

0J

Explanation:

Mass of the lump M1 = 1.0 kg, initial velocity of the lump U1 = 2m/s to the right,

mass of the metal sphere M2 = 0.5kg, initial velocity of the metal sphere M2 = -4m/s to the left. using conservation law of momentum;

M1U1 + M2U2 = V (M1 + M2)

(1.0 × 2) + (0.5 × -4) = V (1 + 0.5)

2 -2 = V (1.5)

0 = V (1.5)

divide both side by 1.5

0/1.5 = V 1.5/1.5

V = 0

since final velocity after sticking together of the two masses equals zero then their combined kinetic energy will also equal to zero

1/2 V(M1 +M2)

1/2 ×0 × 1.5 = 0J

ek = (m·v2) / 2

v - velocity of system after the collision

m = m1 + m2

ek = (m1 + m2) ·v2/2

ek = (200 + 250) ·2,32/2 = 450·5,29/2 = 1190,25 j