91Ó°ÊÓ

The mass of the projectile, \(m_1 = 2.0\,\mathrm{kg}\) both before and during the collision. The initial velocity of the projectile, \(v_{1i} = 8.0 \,\mathrm{m/s}\). The final velocity of the projectile, \(v_{1f} = 6.0\,\mathrm{m/s}\). The angle of deflection, \(\theta = 90.0^{\circ}\). The initial velocity of the target, \(v_{2i} = 0\). Since the collision is elastic we know that: 1. The total linear momentum of the system is conserved. 2. The total kinetic energy of the system is conserved.

The initial momentum of the system, \(p_0\) is the momentum of the projectile since the target is stationary. \(p_0 = m_1 v_{1i}\) The final momentum, \(p_f\) is the resultant momentum of both the projectile and the target after the collision. \(p_f = m_1 v_{1f}^\prime + m_2 v_{2f}^\prime\) Since the angle between the projectile's initial velocity and its final velocity is \(90\) degrees, the final velocities can be resolved into their \(x\) and \(y\) components in the following way: \(v_{1f}^\prime_x = v_{1f} \cos (\theta) = 6.0 * \cos(90^{\circ}) = 0\) \(v_{1f}^\prime_y = v_{1f} \sin (\theta) = 6.0 * \sin(90^{\circ}) = 6.0\) Now, we can write two linear momentum conservation equations: 1. Along the x-axis: \(m_1 v_{1i} = m_1 v_{1f}^\prime_x + m_2 v_{2f}^\prime_x\) 2. Along the y-axis: \(0 = m_1 v_{1f}^\prime_y + m_2 v_{2f}^\prime_y\) Using the above equations, we can solve for \(v_{2f}^\prime_x\) and \(v_{2f}^\prime_y\):

1. Along the x-axis: \(m_1 v_{1i} = m_1 v_{1f}^\prime_x + m_2 v_{2f}^\prime_x\) \(2 * 8.0 = 2 * 0 + m_2 v_{2f}^\prime_x\) \(v_{2f}^\prime_x = \dfrac{2 * 8.0}{m_2}\) 2. Along the y-axis: \(0 = m_1 v_{1f}^\prime_y + m_2 v_{2f}^\prime_y\) \(0 = 2 * 6.0 + m_2 v_{2f}^\prime_y\) \(v_{2f}^\prime_y = - \dfrac{2 * 6.0}{m_2}\)

Now that we have the components of the target body's final velocity, we can find the magnitude of its final velocity. Using the Pythagorean theorem: \(v_{2f} = \sqrt{v_{2f}^{\prime 2}_x + v_{2f}^{\prime 2}_y}\) \(v_{2f} = \sqrt{\left(\dfrac{2 * 8.0}{m_2}\right)^2 + \left(- \dfrac{2 * 6.0}{m_2}\right)^2}\)

Now we can use the conservation of kinetic energy to find \(m_2\): Initial kinetic energy of the system \(= \dfrac{1}{2} m_1 v_{1i}^2\) Final kinetic energy of the system \(= \dfrac{1}{2} m_1 v_{1f}^2 + \dfrac{1}{2} m_2 v_{2f}^2\) Equating the initial and final energies: \(\dfrac{1}{2} m_1 v_{1i}^2 = \dfrac{1}{2} m_1 v_{1f}^2 + \dfrac{1}{2} m_2 v_{2f}^2\) Substituting \(v_{2f}\) from step 4 and solving for \(m_2\): \(\dfrac{1}{2} (2) (8.0)^2 = \dfrac{1}{2} (2) (6.0)^2 + \dfrac{1}{2} m_2 \left(\sqrt{\left(\dfrac{2 * 8.0}{m_2}\right)^2 + \left(- \dfrac{2 * 6.0}{m_2}\right)^2}\right)^2\) Solving for \(m_2\), we get \(m_2 = 2.0\,kg\).

Now that we have the mass \(m_2 = 2.0\,kg\), we can use our expressions for \(v_{2f}^\prime_x\) and \(v_{2f}^\prime_y\) from step 3 to find the components of the target body's final velocity: \(v_{2f}^\prime_x = \dfrac{2 * 8.0}{2} = 8.0\,\mathrm{m/s}\) \(v_{2f}^\prime_y = - \dfrac{2 * 6.0}{2} = -6.0\,\mathrm{m/s}\) Now we simply plug these into the expression for \(v_{2f}\) from step 4: \(v_{2f} = \sqrt{8.0^2 + (-6.0)^2} = 10.0\,\mathrm{m/s}\) So, the speed of the target body after the collision is \(10.0\,\mathrm{m/s}\).