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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