A body of mass 80 kg moving with a velocity of 25 ) ms\(^{-1}\) collides with another moving in the opposite direction at 10 ms\(^{-1}\). After collision, both bodies moved with a common velocity of 12.8 ms\(^{-1}\). Calculate, correct to the nearest whole number, the mass of the second body.
- 33 kg
- 38 kg
-
43 kg
- 47 kg
The correct answer is: C
Explanation
m\(_1\) = 80 kg , m\(_2\) = ?, u\(_1\) = 25 ms\(^{-1}\), u\(_2\) = - 10 ms\(^{-1}\)(opp. direction), v\(_{1, 2}\) = 12.8ms\(^{-1}\)
From the principle of conservation of momentum
m\(_1\) x u\(_1\) + m\(_2\) x u\(_2\) = v\(_{1, 2}\)( m\(_1\) + m\(_2\))
(80 x 25) + (m\(_2\) x (-10)) = 12.8(80 + m\(_2\))
2000 - 10m\(_2\) = 1024 + 12.8m\(_2\)
12.8m\(_2\) + 10m\(_2\) = 2000 - 1024 = 976
22.8m\(_2\) = 976
m\(_2\) = \(\frac{976}{22.8}\) = 42.8 kg ≈ 43 kg to the nearest whole number.