A ball of mass 1.2 kg moving with a velocity | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Mass of balls $m_1 = m_2 = m = 1.2 \ kg$. Initial velocities $u_1 = 12 \ ms^{-1}$,$u_2 = 0$. Coefficient of restitution $e = \frac{1}{\sqrt

Vedclass · Accepted Answer

$3:4$

Vedclass · Answer

(A) Given: Mass of balls $m_1 = m_2 = m = 1.2 \ kg$. Initial velocities $u_1 = 12 \ ms^{-1}$,$u_2 = 0$. Coefficient of restitution $e = \frac{1}{\sqrt{2}}$.By the principle of conservation of linear momentum:$m u_1 + m u_2 = m v_1 + m v_2$$12 + 0 = v_1 + v_2 \Rightarrow v_1 + v_2 = 12 \quad ...(i)$By the definition of coefficient of restitution:$e = \frac{v_2 - v_1}{u_1 - u_2}$$\frac{1}{\sqrt{2}} = \frac{v_2 - v_1}{12 - 0} \Rightarrow v_2 - v_1 = \frac{12}{\sqrt{2}} = 6\sqrt{2} \quad ...(ii)$Adding equations $(i)$ and $(ii)$:$2v_2 = 12 + 6\sqrt{2} \Rightarrow v_2 = 6 + 3\sqrt{2} \ ms^{-1}$Subtracting equation $(ii)$ from $(i)$:$2v_1 = 12 - 6\sqrt{2} \Rightarrow v_1 = 6 - 3\sqrt{2} \ ms^{-1}$Initial kinetic energy $(KE)_i = \frac{1}{2} m u_1^2 = \frac{1}{2} m (12)^2 = 72m$Final kinetic energy $(KE)_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m [(6 - 3\sqrt{2})^2 + (6 + 3\sqrt{2})^2]$$(KE)_f = \frac{1}{2} m [(36 + 18 - 36\sqrt{2}) + (36 + 18 + 36\sqrt{2})] = \frac{1}{2} m [54 + 54] = 54m$Ratio $\frac{(KE)_f}{(KE)_i} = \frac{54m}{72m} = \frac{3}{4}$ or $3:4$.

Similar Questions

In plants,which elements are most readily mobilized?

If $P_1, P_2, P_3$ are the perimeters of the three circles $x^2+y^2+8x-6y=0$,$4x^2+4y^2-4x-12y-186=0$,and $x^2+y^2-6x+6y-9=0$ respectively,then

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

If the charge on the capacitor is increased by $2 \ C$,the energy stored in it increases by $21 \%$. The original charge on the capacitor is: (in $C$)

Observe the following statements:
$(I)$ In $\triangle ABC, b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC, \cot \frac{A}{2} = \frac{b+c}{a} \Rightarrow B = 90^{\circ}$
Which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module