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Correct option is
c) The total amount of momentum before and after the collision is same.
Add some explanation.
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Any idea for question 5?
give the idea of question no.1 also and 5 too thanks
Net force = Tension - Force of gravity = mass * acceleration
a=3.5 and mass=3.50
Is main Kinetic friction kahn gy
Time of flight=1.60s
Time of flight = 2•V•sinΘ/g = 1.60 s ← #1
range = (V²sin(2Θ))/g = 0.75 m ← #2
Divide #2 by #1:
0.75m / 1.6s = Vsin(2Θ) / 2sinΘ
But sin(2Θ)= 2sinΘcosΘ, so
0.47 m/s = VcosΘ ← initial horizontal component -- #3
rearranges to V = 0.47m/s / cosΘ
Now plug that into #1:
2 * (0.47m/s / cosΘ) * sinΘ / 9.8m/s² = 1.6 s
0.096s * tanΘ = 1.6s
tanΘ = 16.7
Now multiply both sides of #3 by tanΘ = 16.7
0.47m/s * 16.7 = VcosΘ * tanΘ = VsinΘ
VsinΘ = 7.8 m/s ← vertical component