Question 199: Centre of Mass and Rotational Motion

Question

In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If
the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative
velocity between the two particles, after collision, is :

Accepted Answer

From conservation of linear momentum,
mv
0
= mv
1
+ mv
2
or v
0
= v
1
+ v
2
........(1)
According to the question,
K
f
= $${3 \over 2}$$K
i
$$ \Rightarrow $$ $${1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} 	imes {1 \over 2}mv_0^2$$
$$ \Rightarrow $$ $$v_1^2 + v_2^2 = {3 \over 2}v_0^2$$
Using eq (1) $${\left( {{v_1} + {v_2}} ight)^2} = v_0^2$$
$$ \Rightarrow $$ $$v_1^2 + v_2^2 + 2{v_1}{v_2}$$ = $$v_0^2$$
$$ \Rightarrow $$ $$2{v_1}{v_2}$$ = $$v_0^2 - {3 \over 2}v_0^2$$ = $$ - {1 \over 2}v_0^2$$
Now, $${\left( {{v_1} - {v_2}} ight)^2}$$ = $${\left( {{v_1} + {v_2}} ight)^2} - 4{v_1}{v_2}$$
= $$v_0^2 - \left( { - v_0^2} ight)$$ = $$2v_0^2$$
$$	herefore$$ $${{v_1} - {v_2}}$$ = $$\sqrt 2 {v_0}$$

Answer

$${{{v_0}} \over {\sqrt 2 }}$$

Answer

$${{v_0}} \over 4$$

Answer

$$\sqrt 2 {v_0}$$

Answer

$${{v_0}} \over 2$$

Centre of Mass and Rotational Motion

In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

Choose the correct answer: