Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
4.8k views
in Physics by (30.0k points)
edited by

A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required. 

1 Answer

+1 vote
by (130k points)
selected by
 
Best answer

Mass of the motor car, m = 1200 kg 

Initial velocity of the motor car, u = 90 km/h = 25 m/s 

Final velocity of the motor car, v = 18 km/h = 5 m/s 

Time taken, t = 4 s 

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

5 = 25 + a (4) 

a = – 5 m/s2 

Negative sign indicates that its a retarding motion i.e. velocity is decreasing. 

Change in momentum = mv – mu = m (v – u)

= 1200 (5 – 25) = – 24000 kg m s -1 

Force = Mass × Acceleration = 1200 × – 5 = – 6000 N 

Acceleration of the motor car = – 5 m/s

Change in momentum of the motor car = – 24000 kg m s -1 

Hence, the force required to decrease the velocity is 6000 N. (Negative sign indicates retardation, decrease in momentum and retarding force)

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

...