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Constants An iron block with mass ma slides down a frictionless hill of height H. At the base of the hill, it collides with and sticks to a magnet with mass mu- • Part A What is the speed e of the block and magnet immediately after the collision? > View Available Hints) O v- ( ) V29H O v= () V2GH O v= () v2gH O v= (m) 29H O v= () 2gH Submit • Part B Now assume that the two masses continue to move at the speed t trom Part A until they encounter a rough surface. The coeficient of friction between the masses and the surface is u. IH the blocks come to rest after a distance s, which of the following eguations would you use to find a? > View Available ints) () aH = umags o gH = umugs o () 9H = p(mp+ mx)gs o ) 9H = -H(ms+ mM)gs o ) gH = p(mp+ mu)9 Submit

Constants An iron block with mass ma slides down a frictionless hill of height H. At the base of the hill, it collides with and sticks to a magnet with mass mu- • Part A What is the speed e of the block and magnet immediately after the collision? > View Available Hints) O v- ( ) V29H O v= () V2GH O v= () v2gH O v= (m) 29H O v= () 2gH Submit • Part B Now assume that the two masses continue to move at the speed t trom Part A until they encounter a rough surface. The coeficient of friction between the masses and the surface is u. IH the blocks come to rest after a distance s, which of the following eguations would you use to find a? > View Available ints) () aH = umags o gH = umugs o () 9H = p(mp+ mx)gs o ) 9H = -H(ms+ mM)gs o ) gH = p(mp+ mu)9 Submit

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Educational Content: Physics Problem Involving Collision and Friction** **Scenario Description:** An iron block with mass \( m_b \) slides down a frictionless hill of height \( H \). At the base of the hill, it collides with and sticks to a magnet with mass \( m_m \). --- **Part A: Collision Analysis** *Question: What is the speed of the block and magnet immediately after the collision?* Available Options: - \( v = \left( \frac{m_b + m_m}{m_b} \right) \sqrt{2gH} \) - \( v = \left( \frac{m_b}{m_b + m_m} \right) \sqrt{2gH} \) - \( v = \left( \frac{m_m}{m_b} \right) \sqrt{2gH} \) - \( v = \left( \frac{m_b}{m_m} \right) \sqrt{2gH} \) - \( v = \sqrt{2gH} \) - \( v = \left( \frac{m_b}{m_b + m_m} \right) 2gH \) --- **Part B: Friction and Motion Analysis** *Question: Now assume that the two masses continue to move at the speed \( v \) from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is \( \mu \). If the blocks come to rest after a distance \( s \), which of the following equations would you use to find \( s \)?* Available Options: - \( \left( \frac{m_b}{m_b + m_m} \right) gH = \mu m_m gs \) - \( \left( \frac{m_b}{m_m} \right) gH = \mu m_b gs \) - \( \left( \frac{m_m}{m_b} \right) gH = \mu (m_b + m_m) gs \) - \( \left( \frac{m_b}{m_m} \right) gH = -\mu (m_b + m_m) gs \) - \( \left( \frac{m_m}{m_b} \right) gH = -\mu m_m gs \) - \( \left( \frac{m_m}{m
Transcribed Image Text:**Educational Content: Physics Problem Involving Collision and Friction** **Scenario Description:** An iron block with mass \( m_b \) slides down a frictionless hill of height \( H \). At the base of the hill, it collides with and sticks to a magnet with mass \( m_m \). --- **Part A: Collision Analysis** *Question: What is the speed of the block and magnet immediately after the collision?* Available Options: - \( v = \left( \frac{m_b + m_m}{m_b} \right) \sqrt{2gH} \) - \( v = \left( \frac{m_b}{m_b + m_m} \right) \sqrt{2gH} \) - \( v = \left( \frac{m_m}{m_b} \right) \sqrt{2gH} \) - \( v = \left( \frac{m_b}{m_m} \right) \sqrt{2gH} \) - \( v = \sqrt{2gH} \) - \( v = \left( \frac{m_b}{m_b + m_m} \right) 2gH \) --- **Part B: Friction and Motion Analysis** *Question: Now assume that the two masses continue to move at the speed \( v \) from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is \( \mu \). If the blocks come to rest after a distance \( s \), which of the following equations would you use to find \( s \)?* Available Options: - \( \left( \frac{m_b}{m_b + m_m} \right) gH = \mu m_m gs \) - \( \left( \frac{m_b}{m_m} \right) gH = \mu m_b gs \) - \( \left( \frac{m_m}{m_b} \right) gH = \mu (m_b + m_m) gs \) - \( \left( \frac{m_b}{m_m} \right) gH = -\mu (m_b + m_m) gs \) - \( \left( \frac{m_m}{m_b} \right) gH = -\mu m_m gs \) - \( \left( \frac{m_m}{m
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