Question

A 4.3 kg object traveling at 2.5 m/s to the right collides head-on with a 3.8 kg object which is at rest. If the impact is perfectly elastic, the velocity of the 3.8 kg object after the impact is ____ m/s.

Answer 1

Answer on Question #69259 Physics / Other

A $m = 4.3 \, kg$ object traveling at $v = 2.5 \, m/s$ to the right collides head-on with a $M = 3.8 \, kg$ object which is at rest. If the impact is perfectly elastic, the velocity of the $3.8 \, kg$ object after the impact is ___ m/s.

Solution:

The momentum conservation law

$$mv = mv' + Mu$$

The energy conservation law

$$\frac{mv^2}{2} = \frac{mv'^2}{2} + \frac{Mu^2}{2}$$

So

$$v' = v - \frac{M}{m} u$$ $$v^2 = \left(v - \frac{M}{m} u\right)^2 + \frac{M}{m} u^2$$ $$v^2 = v^2 - 2 \frac{M}{m} vu + \frac{M^2}{m^2} u^2 + \frac{M}{m} u^2$$ $$u \left(\frac{M^2}{m^2} + \frac{M}{m}\right) = 2 \frac{M}{m} v$$ $$u \left(\frac{M}{m} + 1\right) = 2 v$$ $$u = \frac{2v}{1 + \frac{M}{m}} = \frac{2 \times 2.5}{1 + \frac{3.8}{4.3}} = 2.65 \, \frac{\mathrm{m}}{\mathrm{s}}.$$

Answers: $u = 2.65 \, \frac{\mathrm{m}}{\mathrm{s}}$

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