A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2ms\(^{-1}\). Determine the magnitude of the resulting impulse
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
Impulse = Change in momentum
 m (v - u)
 5 (2.0)
 = 5 x 2 = 10.0kgm\(^{-1}\)
 = 10.0kgms\(^{-1}\)
Impulse = Change in momentum
 m (v - u)
 5 (2.0)
 = 5 x 2 = 10.0kgm\(^{-1}\)
 = 10.0kgms\(^{-1}\)