91Ó°ÊÓ

Since the collision is elastic and friction is absent, the linear momentum is conserved. Let the velocity of the ball after the collision be denoted as \(v_b'\), and the velocity of the wedge after the collision is given as \(4 \mathrm{~m}/\mathrm{s}\). Therefore, we have: \[\text{Initial momentum} = \text{Final momentum}\] \[m_bv_b + m_wv_w = m_bv_b' + m_wv_w'\] Where: \(m_b\) and \(v_b\) are the mass and initial velocity of the ball respectively, \(m_w\) and \(v_w\) are the mass and initial velocity of the wedge respectively, \(v_b'\) is the final velocity of the ball, and \(v_w'\) is the final velocity of the wedge. Plug in the given values: \[1\cdot 10 + 4\cdot 0 = 1\cdot v_b' + 4\cdot 4\] \[10 = v_b' + 16\]

Since the collision is elastic, the coefficient of restitution (e) is equal to 1. Using this, we can write the following equation: \[e = \frac{\text{relative velocity after collision}}{\text{relative velocity before collision}}\] The formula can be written in terms of velocities: \[1 = \frac{(v_w' - v_b') - (v_w - v_b)}{(v_b - v_w) - (v_b' - v_w')}\] Plug in the given values and the relation from the previous step: \[1 = \frac{(4 - v_b') - (0 - 10)}{(10 - 0) - (v_b' - 4)}\]

Now, we'll solve the equation from Step 2 to find the final velocity of the ball, \(v_b'\): \[1 = \frac{14 - v_b'}{10 - v_b' + 4}\] \[1 = \frac{14 - v_b'}{14 - v_b'}\] Since the denominator and numerator of the fraction are equal, and the equation is equal to 1, we conclude that: \[v_b' = 14 - v_b'\] \[v_b' = 7\mathrm{~m}/\mathrm{s}\] The final velocity of the ball (\(v_b'\)) after the collision is \(7 \mathrm{~m}/\mathrm{s}\), which does not match the options given (A, B, C, or D). There might be a mistake, or the options themselves may be incorrect.