A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e = coefficient of restitution, then what will be the ratio of velocity of two balls after collision ?

Question

A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e = coefficient of restitution, then what will be the ratio of velocity of two balls after collision ? (A) (1 -e/1+e) (B) (e -1/e + 1) (C) (1+e/1 -e) (D) (e + 1/e -1)

Tardigrade Admin · Accepted Answer

Here, m1 = m, m2 = m, u1 = u, u2 = 0 Coefficient of restitution, e = (v2 - v1/u1 - u2) e = (v2 -v1/u -0) or v2 - v1 = eu ....(i) According to law of conservation of linear momentum, m1 u1 + m2 u2 = m1 v1 + m2 v2 ⇒ : : mu = m (v1 + v2) or v1 + v2 = u ...(ii) Solving equations (i) and (ii) for v1 and v2, we get v1 = (u(1 -e)/2) , v2 = (u(1 + e)/2) ∴ : : (v1/v2) = (1 -e/1 +e)

Q. A ball of mass $m$ moving with a constant velocity $u$ strikes against a ball of same mass at rest. If $e$ = coefficient of restitution, then what will be the ratio of velocity of two balls after collision ?

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

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

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

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

Q. A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e = coefficient of restitution, then what will be the ratio of velocity of two balls after collision ?

1+e1−e​

e+1e−1​

1−e1+e​

e−1e+1​

Q. A ball of mass $m$ moving with a constant velocity $u$ strikes against a ball of same mass at rest. If $e$ = coefficient of restitution, then what will be the ratio of velocity of two balls after collision ?

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

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

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

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