An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides with and sticks to a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just after collision. *
5 kg m/s
20 kg m/s
15 kg m/s
10 kg m/s
Answers
Given : m
1
=1kg m
2
=5kg u
1
=10 ms
−1
u
2
=0
Momentum of the system just before collision, P
i
=m
1
u
1
+m
2
u
2
P
i
=1×10+5×0=10 kgms
−1
Mass of the combined object, M=m
1
+m
2
=6kg
Let velocity of combined object be v
According to conservation of momentum, momentum of the system just after the collision, P
f
=P
i
⟹P
f
=10 kg m s
−1
Mv=10
or, 6×v=10
or, v=1.67 m/s
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Answer:
10 kg m/s
Explanation:
We have,
m1 = 1 kg
u1 = 10 m/s
m2 = 5 kg
u2 = 0
Also, after collision the velocity of both the object will be same (v).
Using law of conservation of momentum,
Total momentum before collision = Total momentum after collision
⇒ m1*u1 + m2*u2 = (m1+m2)*v
⇒ 1*10 + 5*0 = (1+5)v
⇒ 10 = 6v
⇒ 5/3 = v
Now, total momentum after collision = (m1+m2)*v = 6 kg*5/3 m/s = 10 kg m/s