Two Sumo wrestlers are involved in an inelastic collision. The first wrestler, H

Question

Two Sumo wrestlers are involved in an inelastic collision. The first wrestler, Hakurazan, has a mass of 135 kg and moves forward along the positive x-direction at a speed of 3.5 m/s. The second wrestler, Toyohibiki, has a mass of 173 kg and moves straight toward Hakurazan at a speed of 3.0 m/s. Immediately after the collision, Hakurazan is deflected to his right by 35° (see the figure below). In the collision, 10% of the wrestlers' initial total kinetic energy is lost. What is the angle at which Toyohibiki is moving immediately after the collision?

Explanation / Answer

The initial KE of both are KE= 0.5*M*V2 For Hakurazan Initial KE= 0.5*M*V2 = 0.5*135*3.5*3.5= 826.875 J Final KE= 90percent of initial KE= 90/100 *826.875=744.1875 We can now find the final velocity magnitude of Hakuzaran by the KE formula, v=v(2*KE/M) which gives us Vfh=3.32 m/s Similarly for Toyohibiki, Initial KE=778.5 Final KE=700.65 Hence Vft=v(2*700.65/173) = 2.846 m/s As the net momentum in the Y direction remains 0 both before and after the collision, we can say that Mh Vfh sin(35) = Mt *Vft*sin? solving we get 135*3.32*sin35=173*2.846*sin? 257.076=492.358*sin? sin?=0.52213 ?=31.475 deg

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