Two particles A and B of masses m and 3m are moving along X and Y axes respectively,with the same speed v.They collide at the origin and coalesce into one body,after the collision.What is the velocity of this coalesced mass??
Answers
Answer:
sorry i take 2 instead of 3
Explanation:
Figure shows particles A and B before collision and Figure the coalesced mass after collision.
In figure, we have taken velocity of combinedmass (3m) as V→ making an angle α with positive X-axis.
Using law of conservation of linear momentum
(i) along X-axis
mv=3mVcosα ...(i)
(ii) along Y-axis
2mv=3mVsinα ...(ii)
Dividing (ii) by (i), we get
2=tanαorα=tan−1(2)=63.4∘
Squaring (i) and (ii) and adding, we get
(3mV)2(cos2α+sin2α)=m2v2+m2v2
=5m2v2
V2=59v2orV=5–√3v
No, total K.E.before collision,
K1=12mv2+12(2m)v2=32mv2
total K.E. after collision,
K2=12(3m)V2=12(3m)59v2=56mv2
Loss of K.E. during the collision,
ΔK=K1−K2
ΔK=32mv2−56mv2=23mv2
thanks for your time
Answer:
Linear momentum conservation along X-axis
mv+0=(m+3m) vx
vx=v/4
linear momentum conservation along Y-axis
0+3mv (3m+m) vy
3mv=4m×vy
vy=3v/4
velocity=√vx^2+vy^2
=√10/16v^2
=√5/8×vLinear momentum conservation along X-axis
mv+0=(m+3m) vx
vx=v/4
linear momentum conservation along Y-axis
0+3mv (3m+m) vy
3mv=4m×vy
vy=3v/4
velocity=√vx^2+vy^2
=√10/16v^2
=√5/8×v