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Answer
given,
mass of full back, M = 100 Kg
velocity of full back in east = 3.5 m/s
mass of the defensive back,m = 80 Kg
velocity of defensive back, due east = 6 m/s
a) initial momentum of each player
for full back
P₁ = M v = 100 x 3.5 = 350 kg.m/s
for defensive back
P₂= m v = 80 x 6 = 480 kg.m/s
b) total momentum before collision
P = P₁ + P₂
taking west direction as positive
P = 350 + (- 480 )
P = -130 kg.m/s
c) speed of the them when they stick together.
using conservation of momentum
initial momentum = final momentum
-130 = (M + m ) V
180 V = -130
V = -0.722 m/s
velocity after the collision is equal to 0.722 m/s in direction of east.
a) P₁=350 kg.m/s and P₂= 480 kg.m/s
b) Total momentum before collision is -130 kg.m/s.
c) 0.722 m/s in direction of east
Given,
Mass of full back, M = 100 Kg
Velocity of full back in east = 3.5 m/s
Mass of the defensive back, m = 80 Kg
Velocity of defensive back, due east = 6 m/s
Momentum
It is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.
To find:
a) initial momentum of each player
For full back
P₁ = M*v = 100 * 3.5 = 350 kg.m/s
For defensive back
P₂= m*v = 80*6 = 480 kg.m/s
b) total momentum before collision
P = P₁ + P₂
taking west direction as positive
P = 350 + (- 480 )
P = -130 kg.m/s
c) speed of the them when they stick together.
Using conservation of momentum
initial momentum = final momentum
-130 = (M + m ) V
180 V = -130
V = -0.722 m/s
Velocity after the collision is equal to 0.722 m/s in direction of east.
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brainly.com/question/20182620