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

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`

Step-by-step explanation:

By momentum conservation along x direction we will have

`mv_i = (m)/(2)v_1 + (m)/(2)v_2`

so we have

`v_1 + v_2 = 2v`

`v_1 + v_2 = 4.68`

also by energy conservation

`(1)/(2)((m)/(2))v_1^2 + (1)/(2)((m)/(2))v_2^2 - (1)/(2)mv^2 = 17 J`

`(1)/(4)m(v_1^2 + v_2^2) - (1)/(2)mv^2 = 17`

`(v_1^2 + v_2^2) - 2v^2 = (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 = (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 = (0.24 - 2.34)/(0.16)`

`a_2 = -13.125 m/s^2`