A) 0.19 m
B) 0.379 m
C) 0.569 m
D) 0.758 m
Correct Answer: B
Solution :
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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