Question

A block of mass M with a massless spring of force constant k is resting on a horizontal frictionless surface. A block of mass m projected horizontally with a speed u collides and sticks to the spring at the point of maximum compression of the spring. If v is the velocity of the system after mass m sticks to the spring and n is the fraction of the initial kinetic energy of mass m that is stored in the spring then n is?

Answer 1

Answer on Question # 76285, Physics - Mechanics - Relativity:

Question: A block of mass $M$ with a massless spring of force constant $k$ is resting on a horizontal frictionless surface. A block of mass $m$ projected horizontally with a speed $u$ collides and sticks to the spring at the point of maximum compression of the spring. If $v$ is the velocity of the system after mass $m$ sticks to the spring and $n$ is the fraction of the initial kinetic energy of mass $m$ that is stored in the spring then $n$ is?

Solution: For conservation of momentum and conservation of total energy, we have

$$mu = (M + m)v \quad \text{(1)}$$

And

$$\frac{1}{2}mu^2 = \frac{1}{2}(M + m)v^2 + \frac{1}{2}kx^2 \quad \text{(2)}$$

Dividing equation (2) by equation (1) we get,

$$1 = \frac{(M + m)v^2}{mu^2} + \frac{\frac{1}{2}kx^2}{\frac{1}{2}mu^2}$$

or,

$$\frac{\frac{1}{2}kx^2}{\frac{1}{2}mu^2} = 1 - \frac{(M + m)v^2}{mu^2} \quad \text{(3)}$$

Again from equation (1), we get,

$$\frac{v}{u} = \frac{m}{(M + m)} \quad \text{(4)}$$

Now from equation (3) and equation (4) we get,

$$\frac{\frac{1}{2}kx^2}{\frac{1}{2}mu^2} = \frac{M}{(M + m)} \quad \text{(5)}$$

Answer: $n = \frac{\frac{1}{2}kx^2}{\frac{1}{2}mu^2} = \frac{M}{(M + m)}$

Answer provided by https://www.AssignmentExpert.com