A) 1.8 kg
B) 1.2 kg
C) 1.5 kg
D) 1.0 kg
Correct Answer: B
Solution :
[b] |
By conservation of linear momentum :- |
\[2{{\text{v}}_{0}}=2\left( \frac{{{\text{v}}_{0}}}{4} \right)+m\text{v}\Rightarrow \text{2}{{\text{v}}_{0}}=\frac{{{\text{v}}_{0}}}{2}+m\text{v}\] |
\[\Rightarrow \frac{\text{3}{{\text{v}}_{0}}}{2}=m\text{v}\] .(1) |
Since collision is elastic\[\to \] |
\[{{\text{V}}_{\text{separation}}}\text{=}\,{{\text{v}}_{\text{approch}}}\] |
\[\Rightarrow \text{v-}\frac{{{\text{v}}_{0}}}{4}={{\text{v}}_{0}}\Rightarrow \frac{\text{5}{{\text{v}}_{0}}}{4}=\text{v}\] ..(2) |
equating (2) and (1) |
\[\frac{\text{3}{{\text{v}}_{0}}}{2}=m\left( \frac{\text{5}{{\text{v}}_{0}}}{4} \right)\Rightarrow \,=m=\frac{6}{5}=1.2kg\] |
Option [b] |
You need to login to perform this action.
You will be redirected in
3 sec