Question

Two particles approach each other with equal and opposite speed v. The mass of one particle is m, and the mass of the other particle is nm, where n is just a unitless number. Snapshots of the system before, during, and after the elastic collision are shown above. After the collision the first particle moves in the exact opposite direction with speed 2.40v, and the speed of the second particle is unknown. What is the value of n?

Answer 1

As there is no external force acting on the system, linear momentum conserved. That is total initial momentum is equal

to total final momentum

Apply conservation of energy just berfore and after collision.

$$ \begin{aligned} \frac{1}{2} m v^{2}+\frac{1}{2} n m v^{2} &=\frac{1}{2} m(2.40)^{2} v^{2}+\frac{1}{2} n m v_{n \text { fral }}^{2} \\ v^{2}+n v^{2} &=(2.40)^{2} v^{2}+n v_{n, \text { fral }}^{2} \end{aligned} $$

Rearrange the above equation

$$ \begin{array}{l} n v_{n \text { fral }}^{2}=v^{2}+n v^{2}-(2.40)^{2} v^{2} \\ n v_{\text {n.finl }}^{2}=v^{2}+n v^{2}-(5.76) v^{2} \\ n v_{n, \text { finl }}^{2}=n v^{2}-(4.76) v^{2} \end{array} $$

Substitute value \(\frac{(3.40) v-n v}{n}\) for \(v_{n f \mathrm{mal}}\) \(n\left(\frac{(3.40) v-n v}{n}\right)^{2}=n v^{2}-(4.76) v^{2}\)

\(\frac{(3.40)^{2} v^{2}+n^{2} v^{2}-2 n(3.40) v^{2}}{n}=n v^{2}-(4.76) v^{2}\)

\(\frac{(3.40)^{2} v^{2}+n^{2} v^{2}-2 n(3.40) v^{2}}{n}=n v^{2}-(4.76) v^{2}\)

\((3.40)^{2} v^{2}+n^{2} v^{2}-2 n(3.40) v^{2}=n^{2} v^{2}-n(4.76) v^{2}\)

\((3.40)^{2}+n^{2}-2 n(3.40)=n^{2}-n(4.76)\)

\((3.40)^{2}-2 n(3.40)+n(4.76)=0\)

\(11.56-n(6.80)+n(4.76)=0\)

\(11.56-n(204)=0\)

Solve for \(n\)

$$ n=\frac{11.56}{2.04}=5.66 $$

Final velocity of the second object is,

Negative sign indicates that the object moving al ong negative

\(\mathrm{x}\) -axis.

Speed of the second object after collision is, \(0.40 v\)