Question

Two ice blocks each of mass $6.3$ kg are moving towards each other with $20$m/sec. Initially temperature of each ice block is $0^{0}C$. The head on collision between them is perfectly inelastic. Assume that heat generated due to collision is completely used for melting some quantity of ice. Latent heat of ice for fusion is $80$cal/gm and $1cal=4.2$ Joule. Total mass of water formed due to melting is ($2.5x$) gm then value of $x$ is

• A. 3
• B. 2
• C. 4
• D. 1

Answer 1

The correct option is A 3

Momentum of the system before collision: $P_i$

Momentum of the system after collision: $P_f$

$P_i=P_f$ from momentum conservation

$P_i=0$ since both the blocks have the same initial speed and moving in opposite direction

$\therefore P_f=0$

Since the collision is inelastic, collision both the blocks will combine into one block.

KE is not conserved in an inelastic collision and total KE before collision will get converted into heat.

$4K{E}_{ini}$= 2\times \dfrac{1}{2}mu^2=mu^2$

The heat generated will change the phase of ice to water.

Heat absorbed during phase change from ice to water is $Q=m^{\prime }\times L$

Thus, $m{u}^2=m^{\prime }\times L$ where m' is the amount of ice converted into water during phase change.

$m^{\prime }=\frac{6.3\times {10}^{-3}\times (20{)}^2}{m^{\prime }\times 80\times 4.16}kg$ since 1 cal = 4.16J

$\therefore m^{\prime }=7.5\times {10}^{-3}kg=7.5gm=2.5\times 3gm$

$\therefore x=3$