A 1. 2-kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of 0. 50 m/s. At the origin 0, the bomb explodes into two fragments. Fragment 1 has a mass of 0. 40 kg and a speed of 0. 90 m/s along the negative y-axis. In the figure, the angle θ, made by the velocity vector of fragment 2 and the x-axis, is closest to.



Answer :

By using conservation of linear momentum, it is obtained that the angle [tex]\theta[/tex] is close to 31°

What is conservation of linear momentum?

Conservation of linear momentum states that on collision of two bodies, the total momentum before and after collision remains the same.

Here,

By using conservation of linear momentum

[tex]1.2 \times 0.50 = 0.8 \times vcos\theta\\vcos\theta = \frac{1.2}{0.8}\times 0.50\\vcos\theta = 0.75[/tex]

Also,

[tex]0.4\times 0.9 = 0.8\times vsin\theta\\vsin\theta= \frac{0.4\times 0.9}{0.8}\\vsin\theta = 0.45[/tex]

Now,

[tex]\frac{vsin\theta}{vcos\theta} = \frac{0.45}{0.75}\\tan \theta = 0.6\\\theta = 31^{\circ}[/tex]

To learn about conservation of linear momentum, refer tot the link

https://brainly.com/question/7538238

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