Block A, with a mass of 2 kg, moves along the x axis with a velocity of 5 m/s in

Question

Block A, with a mass of 2 kg, moves along the x axis with a velocity of 5 m/s in the positive x direction. If suffers an elastic collision with block B, initially at rest, and the blocks leave the collision along the x axis. If B is much more massive than A , the velocity of A after the collision is? Block A, with a mass of 2 kg, moves along the x axis with a velocity of 5 m/s in the positive x direction. If suffers an elastic collision with block B, initially at rest, and the blocks leave the collision along the x axis. If B is much more massive than A , the velocity of A after the collision is?

Explanation / Answer

We make use of Law of Conservation of momentum

MaUa+MbUb =MaVa+MbVb

Where Ma and Mb are masses of A and B respectively

U and V represent Initial and Final Velocities.

From conservation of Momentum and kinetic energy we get Va = Ua* [(Ma-Mb)/(Ma+Mb)]

Since Mb >>Ma , we neglect Ma compared to Mb, which gives Va =-Ua

So, Va = - 5 m/s

So, the velocity of A after the collision is -5 m/s

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---