Question

A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting the object into two equal chunks and adding 17 J of kinetic energy to system. The explosive separation takes place over a 0.16-s time interval. Assume that the one of the chunks after explosion moves in the positive x direction. The x component of the average acceleration of this chunk during the explosion is afrontx, the x component fo the average acceleration of the other chunk during the explosion is arearx.
What are the x-components of the average accelerations of the two chunks during the explosion?

Answer 1

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$