A bullet of mass 50 g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g=10ms_(-2)

A bullet of mass 50 g is fired from below into a suspended object of m

1.	A bullet of mass 50 g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g=10ms_(-2)
Answer» Solution :`m_(1)50g=0.05kg,m_(2)=450g=0.45g` The speed of the bullet is up. The second BODY is at rest`(u_(2)=0)`. Let the common velocity of the bullet and the object after the bullet is embedded into the object is V. `v=(m_(1)u_(1)+m_(2)u_(2))/((m_(1)+m_(2))` `v(00.5u_(1)+(0.45xx0))/((00.5+0.45))=(0.05)/(0.50)u_(1)` The combined velocity is the INITIAL velocity for the vertical upward motion of the combine bullet and the object. From second equation of motion, `v=SQRT(2gh)` `v=sqrt(2xx10xx1.8)=sqrt(36)` `v=6ms^(-1)` Substituting this in the above equation, the value of `U_(1)` is `6=(0.05)/(0.50)u_(1) "or" u_(1)(0.50)/(0.05)xx6=10xx6` `v=60ms_(-1)`