A5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at rest.

(a) find the velocity of each object after the collision. cm/s (5.00 g object) cm/s (10.0 g object)
(b) find the fraction of the initial kinetic energy transferred to the 10.0 g object. %

The final speed of the first object is 6.67 cm/s backward and the second object is 13.33 cm/s forward.

The fraction of the initial kinetic energy transferred to the 10.0 g object is 100%.

The final speed of each object is determined by applying the principle of conservation of linear momentum;

Apply one-dimensional velocity;

Solve for the final velocity of the first object;

The final velocity of the second object;

In elastic collision, kinetic energy is conservation. Thus, we can conclude that the fraction of the initial kinetic energy transferred to the 10.0 g object is 100%.

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a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

= 1/2

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