Question

four people a b c and B of masses 795 and 13 in KG respectively are moving with their respective when velocities is in the ratio 4 ratio 3 ratio 5 ratio 2 who will have the highest momentum
option
a -d
B- a
C -b
d- c

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Answer 1

Answer:

The correct option is B-a.

Step-by-step explanation:

Consider the provided information.

It is given that four peoples a, b, c and d of masses 7 Kg, 9 Kg, 5 Kg and 13 Kg.

The ratio of their velocities is 4 : 3 : 5 : 2

Let x is the quantity which multiplied with each term of ratio.

Thus the velocity of the people a, b, c and d is 4x m/s, 3x m/s, 5 x m/s and 2 x m/s respectively.

The formula for Momentum is:

$Momentum(p)= mass \times velocity$

Substitute the respective values.

Momentum of first people= 7×4x=28x Kg m/s

Momentum of second people= 9×3x=27x Kg m/s

Momentum of third people= 5×5x=25x Kg m/s

Momentum of fourth people= 13×2x=26x Kg m/s

The momentum of first people is greatest among them. Hence, the correct option is B-a.