Question

a bullet of mass 0.012 kg and horizontal speed 70 m/e strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block the block suspend3d from the ceiling by two thin wires calculate the height to which the block rises also estimate the amount of heat produced in the block ​

Answer 1

${\green{\underline{\underline{\huge\mathfrak\red{given}}}}}$

m1 = 0.012 kg

u1 = 70 m/s

M2 = 0.4 kg

U2 = 0 m/s

g = 9.8 m/s²

${\green{\underline{\underline{\huge\mathfrak\red{solution}}}}}$

as the bullet strikes the stationary block it comes to rest with respect to the block it gets embedded in the block and thus bullet block system then rises through a height see diagram for reference

let U be the velocity acquired by the system immediately after the perfectly inelastic collision

(I) by the principle of conservation of momentum

(M1 + m2) U = m1u1 + m2u2= m1u1

$u \: = \frac{m1u1}{m1 + m2}$

$= \frac{(0.012kg \: (70 \: ms)}{0.012 + 0.4)kg}$

$= \frac{0.84}{0.412}$

= 2.04 m/s

as the block rises through a height h under the gravitational force

∆PE = - ∆KE

∆(PE f - PE I ) = -(KE f - KEi)

(M1 + m2) gh - 0

$= 0 - \frac{1}{2} (m1 + m2) {u}^{2}$

the bullet block system rises to a height

$h \: = \frac{ {u}^{2} }{2g}$

$= \frac{ {(2.04ms)}^{2} }{ {2(9.8ms)}^{2} }$

$= \frac{4.16}{19.6}$

= 0.2123m

(ii) the kinetic energy of the bullet

$= \frac{1}{2} m1 {u1}^{2}$

$= \frac{1}{2} (0.012kg)( {70ms)}^{2}$

= 0.006 × 4900

= 29.4j

and after the collision with the block the final kinetic energy of the bullet block system

$= \frac{1}{2} {(m1 + m2)}^{2} {u}^{2}$

$= \frac{1}{2} (0.412kg) {(2.04ms)}^{2}$

= 0.206 × 4.162

= 0.8574 j

the loss energy in the perfectly inelastic collision

$\frac{1}{2} m1 {u1}^{2} - \frac{1}{2} (m1 + m2) {u}^{2}$

$= 29.4j \: - 0.8574j$

= 28.54j

the amount of heat produced in the block

$h \: = \frac{loss \: in \: energy}{j}$

where j is the mechanical equivalent of heat taking j

= 4.18 j/cal

$h \: = \frac{28.54}{4.18j \: cal}$

= 6.828

${\green{\underline{\underline{\huge\{\pink{answer \: is \: 6.828 \: cal}}}}}$