Question

A 4000g bomb sliding on a frictionless surface explodes into two 2000g parts, with one having a velocity equal to 3m/s due north, and the other with 5m/s, 30 degrees due north of east, What is the original speed of the bomb?

Answer 1

Assume that y-axis is due North and x-axis is due East:

Apply momentum conservation:

$$x:Mu\text{ cos}\theta=0+mv_2\text{ cos}30°,\\ y: Mu\text{ sin}\theta=mv_1+v_2\text{ sin}30°.$$

Divide second equation by first to find the angle:

$$\theta=\text{atan}\frac{(v_1+v_2\text{ sin}30°)}{v_2\text{ cos}30°}=51.8°$$

(North of East). From the very first equation find the magnitude of the initial velocity:

$$u=\frac{mv_2\text{ cos}30°}{M\text{ cos}\theta}=3.5\text{ m/s}.$$