Physics 410 Exam 3 Fall 2017 (20 points) 3. A 20.0 gram bullet is fired at a 1.4

Question

Physics 410 Exam 3 Fall 2017 (20 points) 3. A 20.0 gram bullet is fired at a 1.48 kg block of wood that is initially at rest and suspended from the top by rod, length L 1.00 m that is able to pivot at the top. The mass of the rod can be ignored). The bullet quickly embeds in the block and the block and bullet swing up to an angle of 40 a. How high does the block swing above its original position? How fast was the block and bullet traveling immediately after impact? What is the speed of the bullet immediately before the collision? b. c.

Explanation / Answer

If mass of rod can be ignored

Conserving momentum

mv=(m+M)*V.......(1)

v is speed of bullet of mass m .V is velocity of bullet and mass after collision

We know mass went upto 40degree

Change in height=l(1-cos(40))=0.2339 m

So

(a) it went 23.38 cm higher

(b) conserving energy we got

V=sqrt(2*9.8*0.2338)=2.14 m/s

(c)using (1) we got

v=160.6 m/s

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