Answer to Question #100850 in Classical Mechanics for Jason

Question

Two wooden pucks approach each other on an ice rink as shown in the figure. Puck #2 has an initial speed of 4.64 m/s and a mass that is some fraction f = 2/3 that of puck #1. Puck #1 is made of a hard wood and puck #2 is made of a very soft wood. As a result, when they collide, puck #1 makes a dent in puck #2 and 12.8% of the initial kinetic energy of the two pucks is lost. Before the collision, the two pucks approach each other in such a manner their momentums are of equal magnitude and opposite directions. Determine the speed of the two pucks after the collision, V1 and V2.

Answer 1

Given that the velocity of the puck 2 =4.64m/sec

and the ratio in the mass f = 2/3

and loss in the energy = 12.8%

As per the question both the puck have the equal momentum,

$P_1=P_2$

$v_1=\dfrac{m_2V_2}{m_2}=\dfrac{3\times 4.64}{2}=6.96m/sec$

$\Rightarrow$ Now lost in energy=$\dfrac{\dfrac{m_1u_1^2+m_2u_2^2}{2}-\dfrac{m_1v_2^2-m_2v_2^2}{2}}{\dfrac{m_1u_1^2+m_2u_2^2}{2}}$

$\Rightarrow \dfrac{12.8}{100}=\dfrac{(16.146+10.76)-(\dfrac{v_1^2+v_2^2}{2})}{(16.146+10.76)}$

$\Rightarrow v_1^2+v_2^2=46.926--------(i)$

The collision is perfectly elastic, so e = 0

$v_1=v_2$

Now substituting the value of $v_1$ in equation (i)

$\Rightarrow 2v_2^2=46.926$

$\Rightarrow v_2^2=\dfrac{46.926}{2}$

$v_2=v_1=\sqrt{23.463}=4.84 m/sec$