A 1.0 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?
1 answer:
Answer:
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
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Answer:
a = 132 m/s^2
Explanation:
f(t) = t^4 − 2t^3 − 6t^2 + 9t
v = f'(t) = 4t^3 - 6t^2 - 12t + 9
a = f''(t) = 12t^2 - 12t -12
Substitute t = 4 seconds into the equation for a:
a = 12(16) -12(4) -12 = 132 m/s^2
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