An object of mass 1.5kg travelling in a straight line with a velocity of 5m/s collides with a wooden block of mass 5kg resting on the floor. This object sticks with the wooden block after collision and both move together in a straight line. Calculate:
(a)the total momentum before collision.
(b) total momentum after collision
(c) the velocity of the combination of these objects after collision
Answers
"Mass of the moving object, m1 = 1.5 kg and its initial velocity, u1 = 5 m/s.
Mass of the wooden block, m2 = 5 kg, and its initial velocity, u2 = 0 m/s.
We need to find total momentum before and after the collision of both objects. Therefore m1u1 + m2u2 = m1v1 + m2v2 => v = 1.15 m/s
Before collision, total momentum 7.5 kgm/s and after collision 7.47 kgm/s."
Answer:
2.25 m/s
Explanation:
Here, m
1
=10g=
1000
10
=0.01kg
m
2
=20g=0.02kg
u
1
=3ms
−1
;u
2
=2ms
−1
v
1
=2.5ms
−1
;v
2
=?
Total momentum of both the spheres before collision =m
1
u
1
+m
2
u
2
=0.01×3+0.02×2=0.07kgms
−1
Total momentum of both the spheres after collision
=m
1
v
1
+m
2
v
2
=0.01×2.5+0.02v
2
=0.025+0.02v
2
Now, according to the law of conservation of momentum,
Total momentum after collision =Total momentum before collision
∴0.025+0.02v
2
=0.07
or 0.02v
2
=0.07−0.025=0.045
or v
2
=
0.02
0.045
=2.25ms
−1