a) A particle of mass mm moving to the right with an initial velocity uu collides elastically with a particle of unknown mass MM coming from the opposite direction. After the collision mm has a velocity u//2u / 2 at right angles to the incident direction, and MM is deflected back making an angle of 45^(@)45^{\circ} degrees to its incident direction as shown below. Calculate the ratio M//mM / m.
Answer:
In an elastic collision, both momentum and kinetic energy are conserved. Let’s consider the horizontal (x) and vertical (y) components of the momentum separately:
Momentum Conservation (Horizontal Direction):
Before the collision, the total horizontal momentum is mu-Mumu – Mu. After the collision, the horizontal momentum of mm is zero (since it moves at right angles to the incident direction), and the horizontal component of MM‘s momentum is Mv cos(45^(@))Mv \cos(45^\circ), where vv is the final velocity of MM. Therefore, we have:
mu-Mu=Mv cos(45^(@))mu – Mu = Mv \cos(45^\circ)
Since cos(45^(@))=(1)/(sqrt2)\cos(45^\circ) = \frac{1}{\sqrt{2}}, this equation simplifies to:
mu-Mu=(Mv)/(sqrt2)mu – Mu = \frac{Mv}{\sqrt{2}}
u(m-M)=(Mv)/(sqrt2)quad(1)u(m – M) = \frac{Mv}{\sqrt{2}} \quad \text{(1)}
Momentum Conservation (Vertical Direction):
Before the collision, the total vertical momentum is zero. After the collision, the vertical momentum of mm is mu//2mu/2, and the vertical component of MM‘s momentum is Mv sin(45^(@))Mv \sin(45^\circ). Therefore:
Kinetic Energy Conservation:
The total kinetic energy before and after the collision remains the same. Before the collision, the total kinetic energy is (1)/(2)mu^(2)+(1)/(2)Mu^(2)\frac{1}{2}mu^2 + \frac{1}{2}Mu^2. After the collision, it is (1)/(2)m(u//2)^(2)+(1)/(2)Mv^(2)\frac{1}{2}m(u/2)^2 + \frac{1}{2}Mv^2. Equating these, we get:
To find the ratio M//mM/m, we can use equations (2) and (3). From equation (2), we have v=(mu)/(Msqrt2)v = \frac{mu}{M\sqrt{2}}. Substituting this into equation (3), we get: