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A block m_(1) strikes a stationary block m_(3) inelastically. Another block m_(2) is kept on m_(3). Neglecting the friction between all contacting surfaces, calculate the fractional decrease in KE of the system in collision. |
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Answer» Solution :Impact takes place between `m_(1)` and `m_(3)` horizontal. Since `m_(2)` is kept on `m_(3)` and all the surface in contact are SMOOTH, friction does not act between `m_(2)` and `m_(3)` during the displacement of `m_(3)` in the impact. Even though there is a friction between `m_(2)` and `m_(3)`, which is very less than the impace FORCE, the frictional force is assumed as non-impulsive force here. Since the impact between `m_(1)` and `m_(2)` is inelastic. `m_(1)` and `m_(3)` will move together toward RIGHT and `m_(2)` will not due to the absence of friction. The VELOCITY of the combined mass `=v^(')=(m_(1)v)/(m_(1)+m_(3))` `(|/_\KE|)/(KE)=(1/2m_(1)v^(2)-1/2(m_(1)+m_(3))v^('2))/(1/2m_(1)v^(2))` `=1-((m_(1)+m_(3))/(m_(1)))((v^('))/v)^(2)` `=1-((m_(1)+m_(3))/(m_(1)))((m_(1))/(m_(1)+m_(3)))^(2)` `=1-(m_(1))/(m_(1)+m_(3))=(m_(3))/(m_(1)+m_(3))` If `m_(2)` and ` m_(3)` are rigidly attached, both together behave as a single mass `(m_(2)+m_(3))` and the answer would have been `(m_(2)+m_(3))/(m_(1)+m_(3)+m_(3))` |
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