Physics: A 5.00 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. -6.67 Correct: Your answer is correct. cm/s (5.00 g object) 13.33 Correct: Your answer is correct. cm/s (10.0 g object) (b) Find the fraction of the initial kinetic energy transferred to the 10.0 g object. 88.44 Correct: Your answer is correct. %

A 5.00 g object moving to the right at 20.0 cm/s

sujeyribetanco2216

18.03.2022

a) $v_{1f}=-6.67\: cm/s$

$v_{2f}=13.33\: cm/s$

b) $n=88.84\: \%$

Explanation:

a) Applying the conservation of momentum, we have:

$p_{i}=p_{f}$

p(i) is the initial momentum. In our case is due to the 5 g object.

p(f) is the final momentum. Here, both objects contribute.

$m_{1i}v_{1i}=m_{1f}v_{1f}+m_{2f}v_{2f}$

Where:

m(1) is 5 gm(1) is 10 gv(1i) is the initial velocity 20 cm/s or 0.2 m/s

To find both final velocities we will need another equation, let's use the conservation of kinetic energy.

$m_{1i}v_{1i}^{2}=m_{1f}v_{1f}^{2}+m_{2f}v_{2f}^{2}$

So we have a system of equations:

$5*0.2=5v_{1f}+10v_{2f}$ (1)

$5*0.2^{2}=5v_{1f}^{2}+10v_{2f}^{2}$ (2)

Solving this system we get:

$v_{1f}=-6.67\: cm/s$

$v_{2f}=13.33\: cm/s$

b) The fraction of the initial kinetic energy transferred is:

$n=\frac{m_{2}v_{2f}^{2}}{m_{1}v_{1i}^{2}}$

$n=\frac{10*13.33^{2}}{5*20^{2}}$

$n=88.84\: \%$

I hope it helps you!

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