Question

An object of mass 3kg travelling in a straight line with a velocity of 5m/s collides with a wooden block of mass 9kg resting on the floor this object gets embedded with a wooden block after collision and both move together in a straight line calculate 1.total Momentum before collision 2.total Momentum after collision 3.velocity of the invented object after collision

Answer 1

total momentum before collision is [3*5 +9*0] = 15 kg m/s
Total momentum after collision =
(5*3+9*0) = (3+9) v
By solving we get v= 1.25m/s total momentum = 150kgm/s
velocity of the invented object is also 1.25 m/s
bcoz it's also embedded in the block

Answer 2

Mass of object =3kg
Mass of wooden block=9kg
According to linear conservation of momentum =m1v1 +m2v2=m1u1 +m2u2
Or total momentum before collision =total momentum after the collision
v(m1+m2)=3*5+9*0
V(3+9)=15
12v=15
V=15/12
V=5/4 =1.25
As total momentum before collision =15 =total momentum after collision