A particle of mass 5kg is kept at the point 2i + 3j + 7k. Another particle of mass 3kg is kept at the point having position vector 5i – 7j –k. Find the position vector of the CoM. If no external force is acting on the system, how would the position of CoM change?
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Answer:
I hope it is helpful
Explanation:
We know that,
L
=m(
r
×
v
)
L
=m[(10
i
^
+6
j
^
)×(5
i
^
)]
L
=m
∣
∣
∣
∣
∣
∣
∣
∣
i
^
10
5
j
^
6
0
k
^
0
0
∣
∣
∣
∣
∣
∣
∣
∣
L
=m[[(10)(0)−(5)(6)]
k
^
+[(6)(0)−(0)(0)]
i
^
+[5(0)−10(0)]
j
^
]
L
=(0.01)[−30
k
^
+(0)
i
^
+(0)
j
^
]
L
=−0.3
k
^
Joule sec
Option A is correct.
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