Particle A mass 0.4kg and B 0.3kg. They move in opposite direction and collide. Before collision, A had speed 6m/s and B had 2m/s. After collision B had 3m/s and moved in opposite direction. Find speed of A after collision with direction and Impulse on B.

-> m1u1 + m2u2 = m1v1 + m2v22.4 - 0.6 = 0.4v1 + 0.90.9 = 0.4 v1v1 = 2.25m/s (direction unchanged)-> I = mv - mu= (0.3 x 3) - (0.3 x -2)= 1.5 Ns

SS
Answered by Samay S. Maths tutor

3880 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

(A) express 4^x in terms of y given that 2^x = y. (B) solve 8(4^x ) – 9(2^x ) + 1 = 0


How do we solve a second order, homogeneous, linear differential equation?


Given y = 2sin(θ) and x = 3cos(θ) find dy/dx.


x = 3t - 4, y = 5 - (6/t), t > 0, find "dy/dx" in terms of t


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning