Answer:
B. 1500 kg*m/s
Explanation:
Momentum p = m* v
In any type of collision, the total momentum is preserved!
The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.
p1 + p2 =
m1*v1 + m2*v2
m1 = me = 300 kg
v1 = 3 m/s
v2 = 2 m/s
Substitute the given numbers:
300*3 + 300+2
900 + 600
1500 kg*m/s, which is answer B.
Answer: 30.34m/s
Explanation:
The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg
Sum of forces in the x direction
mv²/r = N sin 28 + μN cos 28
mv²/r = N(sin 28 + μcos 28)
Thus,
mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]
v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]
v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]
v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]
v²/36 = 9.8 (1.2376/0.4744)
v²/36 = 9.8 * 2.6088
v²/36 = 25.57
v² = 920.52
v = 30.34m/s
Answer:
0.1 m
Explanation:
F = Force exerted on spring = 3 N
k = Spring constant = 60 N/m
x = Displacement of the block
As the energy of the system is conserved we have
The position of the block is 0.1 from the initial position.
Answer:
7.74m/s
Explanation:
Mass = 35.9g = 0.0359kg
A = 39.5cm = 0.395m
K = 18.4N/m
At equilibrium position, there's total conservation of energy.
Total energy = kinetic energy + potential energy
Total Energy = K.E + P.E
½KA² = ½mv² + ½kx²
½KA² = ½(mv² + kx²)
KA² = mv² + kx²
Collect like terms
KA² - Kx² = mv²
K(A² - x²) = mv²
V² = k/m (A² - x²)
V = √(K/m (A² - x²) )
note x = ½A
V = √(k/m (A² - (½A)²)
V = √(k/m (A² - A²/4))
Resolve the fraction between A.
V = √(¾. K/m. A² )
V = √(¾ * (18.4/0.0359)*(0.395)²)
V = √(0.75 * 512.53 * 0.156)
V = √(59.966)
V = 7.74m/s
