The 240-g projectile is fired with a velocity of 930 m/s towards the center of t

Question

The 240-g projectile is fired with a velocity of 930 m/s towards the center of the 12-kg wooden block, which rests on a rough surface. If the projectile penetrates and emerges from the block with a velocity of 360 m/s, determine the velocity of the block just after the projectile emerges. How long(time & distance) does the block slide on the rough surfaces, after the projectile emerges, before it comes to rest again? The coefficient of kinetic friction between the surface and the block is mk = 0.25.

Explanation / Answer

by conserving linear momentum we get,(024)*930=(0.24)*360+12*v........=>v=11.4m/s.......therefore,the velocity with which the block moves is 11.4m/s....now,taking the frictional force into consideration, F=mk(mg)=29.4N...hence deceleration=29.4/12=2.45m/s^2 ....now ,applying the formula v^2-u^2=2as we get distance = 26.52m and by applying v=u+at we get t=4.65s....plz rate :)

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