Fewpal
+1 vote
276 views
in Physics by (29.7k points)
edited by

A bullet of mass 10 g travelling horizontally with a velocity of 150 ms-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet. 

1 Answer

+1 vote
by (127k points)
selected by
 
Best answer

Now, it is given that the bullet is travelling with a velocity of 150 m/s. 

Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s 

Final velocity, v = 0 (since the bullet finally comes to rest) 

Time taken to come to rest, t= 0.03 s 

According to the first equation of motion, v = u + at 

Acceleration of the bullet, a 

0 = 150 + (a × 0.03 s)

(Negative sign indicates that the velocity of the bullet is decreasing.) 

According to the third equation of motion: 

v2= u2 + 2as 

0 = (150)2 + 2 ( - 5000) s 

Hence, the distance of penetration of the bullet into the block is 2.25 m. 

From Newton's second law of motion: 

Force, F = Mass x Acceleration 

Mass of the bullet, m = 10 g = 0.01 kg 

Acceleration of the bullet, a = 5000 m/s

F = ma = 0.01×5000 = 50 N

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...