Answer:
41.27m/s
Explanation:
According to law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the velocity after impact
Given
m1 = 0.2kg
u1 = 43.7m/s
m2 = 45.9g = 0.0459kg
u2 = 30.7m/s
Required
Velocity after impact v
Substitute the given parameters into the formula
0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v
8.74+1.409 = 0.2459v
10.149 = 0.2459v
v = 10.149/0.2459
v = 41.27m/s
Hence the speed of the golf ball immediately after impact is 41.27m/s
In this problem, we use moment balance to get how far the boat moves. In this case, we multiply the mass of an object multiplied by the distance of the object. In this case, the balance goes:
1326 kg * 6.5 m = 1.2 x 104 kg * x where x is the distance
x is equal to 0.71825 meters.
Vi = Velocity initial = 35 m/s
Vf = Velocity final = 75 m/s
t = 5s
Equation:
Vf = Vi + at
65 = 35 + a(5)
30 = 5a
a = 6
Acceleration was 6 m/s^2
C. 6 m/s^2
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