JAMB - Physics (2018 - No. 74)
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
20.0kgms\(^{-1}\)
10.0kgms\(^{-1}\)
5.0kgms\(^{-1}\)
2.5kgms\(^{-1}\)
Wyjaśnienie
Impulse = Change in momentum
m (v - u)
5 (2.0)
= 5 x 2 = 10.0kgm\(^{-1}\)
= 10.0kgms\(^{-1}\)