Question on: JAMB Physics - 2018
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
A
20.0kgms\(^{-1}\)
B
10.0kgms\(^{-1}\)
C
5.0kgms\(^{-1}\)
D
2.5kgms\(^{-1}\)
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Correct Option: B
Impulse = Change in momentum
m (v - u)
5 (2.0)
= 5 x 2 = 10.0kgm\(^{-1}\)
= 10.0kgms\(^{-1}\)
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