Answer to Question #273973 in Quantum Mechanics for Reynilda A Guindol

Question

Suppose two lumps of clay approach each other from opposite

directions in deep space. Lump number 1 masses 4.00 kg and travels at

5.00 m/s. Lump number 2 masses 10.0 kg. When the lumps collide, no

material flies off. The lumps stick together , and the resulting object is

stationary relative to you, the observer. How fast was lump number 2

moving before the collision relative to you, the observer?

Answer 1

Let us calculate the momenta of lumps. Before the collision the first lump had a momentum of $m_1v_1 = 4.00\,\mathrm{kg}\cdot5.00\mathrm{m/s}=20\,\mathrm{kg\cdot m/s}.$

The second momentum is $m_2v_2 = 10.0\,\mathrm{kg}\cdot v_2$

and the direction of this momentum is opposite to the direction of the first momentum.

Due to the law of conservation of momenta we have

$m_1v_1 - m_2v_2 =(m_1+m_2)v,$

where we take the opposite directions of motion into account and where $v=0$.

Therefore,

$m_1v_1=m_2v_2, v_2 = 2\,\text{m/s}.$