An object initially at rest suddenly explodes in two fragments of masses $2.6 \mathrm {~kg}$ and $3.3 \mathrm {~kg}$ that move in opposite directions. If the speed of the first fragment is $3.6 \mathrm {~m} / \mathrm { s } \text {, }$ find the internal energy of the explosion.
A) 30 J
B) 30 kJ
C) 38 J
D) 38 kJ

Question

Quizplus · Accepted Answer

The internal energy of the explosion is equal to the kinetic energy of the fragments. The kinetic energy of the first fragment is:$$K_1 = \frac{1}{2}mv_1^2 = \frac{1}{2}(2.6 \mathrm{~kg})(3.6 \mathrm{~m/s})^2 = 18.48 \mathrm{~J}$$The kinetic energy of the second fragment is:$$K_2 = \frac{1}{2}mv_2^2 = \frac{1}{2}(3.3 \mathrm{~kg})v_2^2$$Since the fragments move in opposite directions, the total kinetic energy is:$$K = K_1 + K_2 = 18.48 \mathrm{~J} + \frac{1}{2}(3.3 \mathrm{~kg})v_2^2$$The internal energy of the explosion is equal to the total kinetic energy, so:$$U = K = 18.48 \mathrm{~J} + \frac{1}{2}(3.3 \mathrm{~kg})v_2^2$$We don't know the value of $v_2$, but we can use the conservation of momentum to find it. The total momentum of the system before the explosion is zero, so the total momentum after the explosion must also be zero:$$0 = (2.6 \mathrm{~kg})(3.6 \mathrm{~m/s}) + (3.3 \mathrm{~kg})v_2$$Solving for $v_2$, we get:$$v_2 = -2.7 \mathrm{~m/s}$$Substituting this value into the equation for the internal energy, we get:$$U = 18.48 \mathrm{~J} + \frac{1}{2}(3.3 \mathrm{~kg})(-2.7 \mathrm{~m/s})^2 = 30 \mathrm{~J}$$Therefore, the internal energy of the explosion is 30 J.

An Object Initially at Rest Suddenly Explodes in Two Fragments 2.6 kg2.6 \mathrm {~kg}2.6 kg

An Object Initially at Rest Suddenly Explodes in Two Fragments $2.6 \mathrm {~kg}$