An object of mass 1 kg travelling in a straight line with a velocity of 10m/s^-1 collides with, & sticks to, a stationary wooden block of mass 5kg. Then they both move of together in the same straight line. Calculate the total momentum just before the impact & just after the impact.Also, calculate the velocity of the combined object.
Answers
Step-by-step explanation:
By law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
As,we have to find combined velocity of the system,therefore,take v1 = v2 = V,then law of conservation of momentum becomes:-
m1u1 + m2u2 = (m1 + m2)V - eq 1
Here,m1 = mass of object = 1kg
m2=mass of stationary wooden block=5kg
u1=velocity of object before collision/initial velocity of object = 10m/s
u2=velocity of stationary wooden stick before collision = 0m/s(because,block is stationary).
V = ?
Put the above mentioned values in eq - 1
=>1×10 + 5×0 = (1+5)V
=>10+0=6V
=>10/6 = V
=>V= 1.67m/s = velocity of combined object
Total momentum just before the impact = m1u1 + m2u2 = 10kgm/s
Total momentum just after collision = (m1+m2)V = 10.02kgm/s = 10kgm/s (approx.)