An object located at the origin and having mass \(M\) explodes into three pieces having masses \(M / 4, M / 3,\) and \(5 M / 12 .\) The pieces scatter on a horizontal frictionless xy-plane. The piece with mass \(M / 4\) flies away with velocity \(5.0 \mathrm{m} / \mathrm{s}\) at \(37^{\circ}\) above the \(x\)-axis. The piece with mass \(M / 3\) has velocity \(4.0 \mathrm{m} / \mathrm{s}\) directed at an angle of \(45^{\circ}\) above the \(-x\)-axis. (a) What are the velocity components of the third piece? (b) Describe the motion of the CM of the system after the explosion.

To find the momenta of the two masses, we will first calculate their velocity components in the x and y directions. We can use the following equations to determine these components: Velocity component in x-direction: \(v_x = v * \cos(\theta)\) Velocity component in y-direction: \(v_y = v * \sin(\theta)\) Now, let's find the momenta for the piece with mass \(M/4\) at an angle of \(\theta_1 = 37^\circ\). \(v_{1x} = 5.0 * \cos(37^\circ) = 4 \, m/s\) \(v_{1y} = 5.0 * \sin(37^\circ) = 3 \, m/s\) \(m_1 = M/4\) \(p_{1x} = m_1v_{1x} = (M/4)(4) = M \, kg*m/s\) \(p_{1y} = m_1v_{1y} = (M/4)(3) = (3/4)M \, kg*m/s\) Similarly, we find the momenta for the piece with mass \(M/3\) at an angle of \(\theta_2 = 45^{\circ}\) above the negative x-axis. \(v_{2x} = -4.0 * \cos(45^\circ) = -2\sqrt{2} \, m/s\) \(v_{2y} = 4.0 * \sin(45^\circ) = 2\sqrt{2} \, m/s\) \(m_2 = M/3\) \(p_{2x} = m_2v_{2x} = (M/3)(-2\sqrt{2}) = -(2/3)\sqrt{2}M \, kg*m/s\) \(p_{2y} = m_2v_{2y} = (M/3)(2\sqrt{2}) = (2/3)\sqrt{2}M \, kg*m/s\)

Using the conservation of momentum, we can find the momentum components of the third piece. The total momentum before the explosion is zero. Hence, \(p_{3x} = -p_{1x} - p_{2x} = -M + \frac{2}{3} \sqrt{2}M\) \(p_{3y} = -p_{1y} - p_{2y} = -\frac{3}{4}M - \frac{2}{3}\sqrt{2}M\)

Next, we will find the velocity components of the third piece with mass \(m_3 = 5M/12\). We can do this by dividing the momentum components by the mass: \(v_{3x} = \frac{p_{3x}}{m_3} = \frac{-M + \frac{2}{3} \sqrt{2}M}{5M/12} = -\frac{8}{5} + \frac{4\sqrt{2}}{5} \,m/s\) \(v_{3y} = \frac{p_{3y}}{m_3} = \frac{-\frac{3}{4}M - \frac{2}{3}\sqrt{2}M}{5M/12} = -\frac{9}{5} - \frac{8\sqrt{2}}{15} \,m/s\) Therefore, the velocity components of the third piece are \((-\frac{8}{5} + \frac{4\sqrt{2}}{5}) \,m/s\) in the x-direction and \((-\frac{9}{5} - \frac{8\sqrt{2}}{15}) \,m/s\) in the y-direction.

To describe the motion of the center of mass (CM) of the system after the explosion, we will calculate the position of the CM (using the masses and the velocities found earlier). We can compute the positions using the following equations: \(x_{CM} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}\) \(y_{CM} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}\) However, the total momentum of the system after the explosion is still zero (since the initial momentum was zero). That means the position of the center of mass will remain at the same point where the object initially exploded (the origin). Thus, the motion of the CM of the system after the explosion remains at the origin, and it doesn't move.