Answer to Question #181239 in Mechanics | Relativity for erick

Question

he two balls in the figure above have an elastic collision.

Before the collision ball A was moving at 4.1 m s⁻¹ and ball B was stationary. After the collision ball A is moving at 2.5 m s⁻¹.

What is the speed of ball B after the collision? (in m s⁻¹ to 2 s.f)

Answer 1

Before collision,

$v_a=4.1ms^{-1}, v_b=0$

After collision,

$v_{a'}=2.5ms^{-1}$

using the formula of coefficinet of restitution-

$e=\dfrac{v_{b'}-v_{a'}}{v_b-v_a}\Rightarrow1=\dfrac{v_{b'}-2.5}{0-4.1}$

$\Rightarrow -0.41=v_{b'}-2.5$

$v_{b'}=2.5-0.41=2.09ms^{-1}$

Hence The velocity of Ball B after collision is $2.09ms^{-1}$