Question

A bullet of mass 0.012 kg and horizontal speed `70ms^(-1)` strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by thin wire. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer 1

Here, `m_(1)= 0.012kg, u_(1)= 70m//s`.
`m_(2)= 0.4 kg,` `u_(2)=0`
As the bullet comes to rest with respect to the block , the two behave as one body. Let v be the velocity acquired by the combinatin.
Applying principle of conservation of linear momentum, `(m1+m2)v=m1H1 + m2u2=m1u1`
`v=(m_(1)u_(1))/(m_(1)+m_(2)) = (0.012 xx 70)/(0.012 +0.4) = 0.84/0.412 = 2.04 ms^(-1)`
Let the block rise to a height h.
P.E. of the combination = K.E of the combinatiion
`(m_(1) + m_(2))gh= 1/2(m_(1)+m_(2))v^(2)` ltbgt `therefore h=v^(2)/(2g) = (2.04 xx 2.04)/(2 xx 9.8) = 0.212 m`
For calculating heat produced, we calculate energy lost (W), where W=intial K.E of bullet - final K.E. of combination.
`=1/2m_(1)u_(1)^(2)-1/2(m_(1)+m_(2))v^(2)`
`1/2 xx 0.012 (70)^92)-1/2(0.412)(2.04)^(2)`
W`=29.4 - 0.86 = 28.54` Joule
`therefore` Heat produced , `H=W/J = 28.54/4.2 = 6.8 cal`