A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass. If e is the coefficient of restitution, then the ratio of velocities of the two spheres after collision ((v1/v2)) will be

Question

A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass. If e is the coefficient of restitution, then the ratio of velocities of the two spheres after collision ((v1/v2)) will be (A) (1-e/1+e) (B) (1+e/1-e) (C) (e+1/1-e) (D) (e-1/e+1)

Tardigrade Admin · Accepted Answer

Conservation of momentum (gives) m u+0=m v1+m v2 ⇒ v1+v2=u ...(i) Newton's experimental formula, v1-v2=-e[u1-u2] gives v1-v2=-e u ...(ii) From Eqs. (i) and (ii), we get v1=((1-e) u/2) v2=((1+e) u/2) ∴ (v1/v2)=(1-e/1+e)

Q. A sphere of mass $m$ moving with a constant velocity $u$ hits another stationary sphere of the same mass. If $e$ is the coefficient of restitution, then the ratio of velocities of the two spheres after collision $(\frac{v _{1}}{v _{2}})$ will be

$\frac{1 - e}{1 + e}$

$\frac{1 + e}{1 - e}$

$\frac{e + 1}{1 - e}$

$\frac{e - 1}{e + 1}$

Q. A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass. If e is the coefficient of restitution, then the ratio of velocities of the two spheres after collision (v2​v1​​) will be

1+e1−e​

1−e1+e​

1−ee+1​

e+1e−1​

Conservation of momentum (gives) mu+0=mv1​+mv2​ ⇒v1​+v2​=u ...(i) Newton's experimental formula, v1​−v2​=−e[u1​−u2​] gives v1​−v2​=−eu ...(ii) From Eqs. (i) and (ii), we get v1​=2(1−e)u​ v2​=2(1+e)u​ ∴v2​v1​​=1+e1−e​

Q. A sphere of mass $m$ moving with a constant velocity $u$ hits another stationary sphere of the same mass. If $e$ is the coefficient of restitution, then the ratio of velocities of the two spheres after collision $(\frac{v _{1}}{v _{2}})$ will be

$\frac{1 - e}{1 + e}$

$\frac{1 + e}{1 - e}$

$\frac{e + 1}{1 - e}$

$\frac{e - 1}{e + 1}$