Question

A 128g block sits at rest on a level surface. There is a coefficient of kinetic friction between the block and the surface of H = 0.500. An 8.00g bullet is fired into the block at vo= 300 other side of the block travelling at a speed of v 140. No mass is lost during the collision and the and emerges from the block moves in the same direction as the bullet after the collision tX (i) no 1esv (f) Mou 8:00g-mau My=0.500 v mo1209 m What is the speed, u, of the block immediately after the collision? Indicate + above. [8pts) a. on the diagram

Answer 1

Here we apply momentum conservation of system and find the final velocity of block just after impact. Then apply Newton's law to find deceleration of block. Using kinematics equation we find distance travelled by block.

Mcus Black M A SDo - Maw gm0.008tg al bullest jial vedoclty af black y = V velacity afblock net externalfore on (bullet VD. so linear momenten ll te M VD ML fA F3AW (6.008)(300) (o 008(14o) Co V 4peet blachpat after coOn tward inpulae Chanye inmamentn Dmpulae impadel block y bulet MV O. 128 ihtad CCollisan ie 2S de c We

125 Ithecaseinidtialseees will e the peed afblactjtaftex calliaion Now S= fret Ma = M CA M Or Let blochmaer datance X to reut 2aXtu final velocity 20592) +CO) lo.2meter