Question

Sphere A has mass m and is moving with velocity v. It makes a head-on elastic collision with a stationary sphere B of mass 2m. After the collision their speeds (vA and vB) are:

Answer 1

@michele_laino

Answer 2

@irishboy123

Answer 3

Conserve momentum and also energy.

Initial total momentum and energy would be
mv and 0.5mv^2

Final total momentum and energy would be

mvA + 2mvB and 0.5mvA^2+ 0.5(2m)vB^2

Answer 4

Hint: \(\sf V_B = \Large \frac{2M_1V}{M_1+M_2}\)

Answer 5

@mashy and @abhisar it would be very kind of you to lead me to the conclusion and get me the final answer....

Answer 6

step 1
formally write out the equation for conservation of momentum as it applies here

do that and we can go to step 2

Answer 7

since we have an elastic collision, and such collision happen along a horizontal line, then we have to apply the conservation of momentum (our mechanical system is isolated) and the conservation of kinetic energy ( there is no potential energy of interaction), so we can write this:
\[\large \left\{ \begin{gathered}
mv = m{u_1} + \left( {2m} \right){u_2}\quad \left( {{\text{conservation of momentum}}} \right) \hfill \\
\hfill \\
\frac{{m{v^2}}}{2} = \frac{{mu_1^2}}{2} + \frac{{\left( {2m} \right)u_2^2}}{2}\quad \left( {{\text{conservation of kinetic energy}}} \right) \hfill \\
\end{gathered} \right.\]
where \( \large u_1, \; u_2 \) are the speed after collision, and \( \large v \) is the speed of the sphere \( \large A \) before collision.
Please solve that system for \( \large u_1, \; u_2\)