A ball of mass M moving with a speed of 2 m | vedclass

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(C) The velocity of the center of mass $(v_{cm})$ is given by: $v_{cm} = \frac{M \cdot v_1 + m \cdot v_2}{M + m} = \frac{M(2) + 1(1)}{M + 1} = \frac{2

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(C) The velocity of the center of mass $(v_{cm})$ is given by: $v_{cm} = \frac{M \cdot v_1 + m \cdot v_2}{M + m} = \frac{M(2) + 1(1)}{M + 1} = \frac{2M + 1}{M + 1}$.The kinetic energy of the center of mass is given by: $K.E._{cm} = \frac{1}{2} (M + m) v_{cm}^2$.Given $K.E._{cm} = \frac{4}{3} \ J$,we substitute the values:$\frac{4}{3} = \frac{1}{2} (M + 1) \left( \frac{2M + 1}{M + 1} \right)^2$.$\frac{4}{3} = \frac{1}{2} \frac{(2M + 1)^2}{M + 1}$.$8(M + 1) = 3(4M^2 + 4M + 1)$.$8M + 8 = 12M^2 + 12M + 3$.$12M^2 + 4M - 5 = 0$.Solving the quadratic equation using the formula $M = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$M = \frac{-4 \pm \sqrt{16 - 4(12)(-5)}}{2(12)} = \frac{-4 \pm \sqrt{16 + 240}}{24} = \frac{-4 \pm \sqrt{256}}{24} = \frac{-4 \pm 16}{24}$.Since mass must be positive,$M = \frac{12}{24} = 0.5 \ kg$.

$A$ ball of mass $M$ moving with a speed of $2 \ m \ s^{-1}$ hits another ball of mass $1 \ kg$ moving in the same direction with a speed of $1 \ m \ s^{-1}$. If the kinetic energy of the center of mass is $4/3 \ J$,then the magnitude of $M$ is: (in $kg$)

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