# Thread: Inelastic collision of right angled bodies

1. ## Inelastic collision of right angled bodies

1. The problem statement, all variables and given/known data
Hi
Can someone help with this problem.
2 particles of mass 30g and 40g respectively both travel at a speed of 35m/s in directions at right angles. The 2 particles collide and stick together. Calculate their speed after impact.

2. Relevant equations
MaVa + MbVb =(Ma + Mb)v
v= common velocity

3. The attempt at a solution
I am not sure how to start. Solving the total momentum before and total momentum after gives the original speed. I then resolved the respective velocities. But that means I got my final value without consideration of the masses! Finally, I resolved the masses into forces, but still didnt get anywhere. Where I am going wrong? Please advice

2. Originally Posted by gbenguse78
1. The problem statement, all variables and given/known data
Hi
Can someone help with this problem.
2 particles of mass 30g and 40g respectively both travel at a speed of 35m/s in directions at right angles. The 2 particles collide and stick together. Calculate their speed after impact.

2. Relevant equations
MaVa + MbVb =(Ma + Mb)v
v= common velocity

3. The attempt at a solution
I am not sure how to start. Solving the total momentum before and total momentum after gives the original speed. I then resolved the respective velocities. But that means I got my final value without consideration of the masses! Finally, I resolved the masses into forces, but still didnt get anywhere. Where I am going wrong? Please advice
Since we are dealing with a two dimensional collision, we need to use the following formulas:

\displaystyle \begin{aligned}m_1v_{1x_i}+m_2v_{2x_i}&=(m_1+m_2)v _x\\m_1v_{1y_i}+m_2v_{2y_i}&=(m_1+m_2)v_y\end{alig ned}

Without loss of generality, I'm going to assume the following:

* Each mass moves in the positive axis directions -- $\displaystyle m_1=.03\,kg$ in the positive x direction, and $\displaystyle m_2=.04\,kg$ in the positive y direction.

So, with this information, we have the following:

\displaystyle \begin{aligned} (.03\,kg)(35\,m/s) &= (.03\,kg+.04\,kg)v_x\\ (.04\,kg)(35\,m/s) &= (.03\,kg+.04\,kg)v_y \end{aligned}

With this, you should be able to find the x and y components of your velocity.

Finally, your answer will be $\displaystyle v=\sqrt{v_x^2+v_y^2}$.

Does this make sense? Can you finish this off?

3. Yes! Thanks.