A) 4.9 m/s upward
B) 4.9 m/s downward
C) 9.8 m/s upward
D) 9.8 m/s downward
Correct Answer: A
Solution :
Let balls collide at time t, they are at same height h. Height of first ball after t seconds \[=49\,t-0.5\,(9.8\,{{t}^{2}})=4.9\,t\,(10-t)\] Height of second ball after t seconds = 98 - downward distance moved by it in t seconds \[=98-0.5(9.8)\,{{t}^{2}}=4.9\,(20-{{t}^{2}})\] \[\Rightarrow \] \[4.9\,(10\,-t)=4.9\,(20-{{t}^{2}})\] \[\Rightarrow \] \[10\,t-{{t}^{2}}=20-{{t}^{2}}\] \[\Rightarrow \] \[t=2\,s\] The balls thus collide after 2 s the start of their motion. Their velocities at this instance are Ball 1 : \[{{v}_{1}}=(~49-98\times 2)=29.4\text{ }m/s\] upward Ball 2 : \[{{v}_{2}}=(0+9.8\times 2)=19.6\text{ }m/s\] downwards From conservation of momentum \[200\,v=100\times 29.4-100\times 19.6\]\[\Rightarrow \]\[v=4.9\,m/s\]
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