wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The initial speed of a body of mass 2.0 kg is 5.0 ms1. A force acts for 4 seconds in the direction of motion of the body. The force-time graph is shown in figure. Calculate the impulse of the force and also the final speed of the body.

A
8.50 N s,9.25 ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4.25 N s,9.25 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4.25 N s,4.25 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8.50 N s,4.25 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 8.50 N s,9.25 ms1
The magnitude of the impulse of a force is equal to the area between the force-time curve and the time-axis.
Impulse of given force i.e., J= area of oabcde
J= area of triangle oaa + area of reactangle aabb + area of trapezium bbcc + area of rectangle cced
=12×1.5×3+1×3+12(3+2)0.5+2×1
=2.25+3+1.25+2=8.50 N s
Impulse = change in momentum (mΔv)
Where Δv is the increase in velocity.
Δv=Impulsem=8.502 ms1=4.25 ms1
Final speed = Initial speed + Δv
Final speed 00000=5.00 ms1+4.25 ms1=9.25 ms1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Second Law of Motion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon