A) 0.19 m
B) 0.379 m
C) 0.569 m
D) 0.758 m
Correct Answer: B
Solution :
| [b] Given, \[{{m}_{1}}=4g,{{u}_{1}}=300m/s\] |
| \[{{m}_{2}}=0.8kg=800g,{{u}_{2}}=0m/s\] |
| From law of conservation of momentum, |
| \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] |
| Let the velocity of combined system = v m/s |
| then,\[4\times 300+800\times 0=(800+4)\times v\] |
| \[v=\frac{1200}{804}=1.49m/s\] |
| Now, \[\mu =0.3\] (given) |
| \[a=\mu g\] |
| \[a=0.3\times 10\] (take \[g=10\,m/{{s}^{2}}\]) |
| \[=3m/{{s}^{2}}\]then, from \[{{v}^{2}}={{u}^{2}}+2\]as |
| \[{{(1.49)}^{2}}=0+2\times 3\times s\]\[s=\frac{{{\left( 1.49 \right)}^{2}}}{6}\] |
| \[s=\frac{2.22}{6}\] |
| \[=0.379m\] |
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