Test Index

JAMB Physics 2021 Paper

Question : 21

Total: 40

^{-1}

Solution:

Given

_{1}

= 0.05kg, u

_{1}

= 200ms

^{-1}

, m

_{2}

= 0.95kg

K.E = $\frac{1}{2}$ m $_{r}$ v $^{2}$

m $_{1}$ u $_{1}$ = v(m $_{1}$ + m $_{2}$ ) [law of conversation of momentum]

v = $\frac{0.05 \times 200}{0.05 + 95}$ = 10ms $^{-1}$

K.E = $\frac{1}{2}$ (1)10 $^{2}$ = 50J

Go to Question:

Prev Question

Next Question