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maria [59]
1 year ago
10

Two carts are free to slide along the frictionless track shown in figure below. Cart A of mass m1 = 8 kg is released from 12m. A

cart B of mass m2 = 4 kg, initially at rest. The two carts combine together and move as one object. Calculate the height reached
by the two carts after collision.
Physics
1 answer:
Harrizon [31]1 year ago
3 0

The height reached by the two carts after collision is determined as 5.34 m.

<h3>Initial velocity of Cart A</h3>

Apply the principle of conservation of mechanical energy.

K.E = P.E

v = √2gh

v = √(2 x 9.8 x 12)

v = 15.34 m/s

<h3>Final velocity of the two carts after the collision</h3>

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

8(15.34) + 4(0) = v(8 + 4)

122.72 = 12v

v = 10.23 m/s

<h3>Height reached by both carts</h3>

Apply the principle of conservation of mechanical energy.

P.E = K.E

mgh = ¹/₂mv²

h = v²/(2g)

h = (10.23²) / (2 x 9.8)

h = 5.34 m

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

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Answer:

E=hf

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4 0
2 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
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(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
Read 2 more answers
. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from th
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Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

F \Delta t = m (v-u)

where

F is the average force

\Delta t is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

\Delta t = 60.0 ms = 0.06 s

Solving for F,

F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N

And since we are interested in the magnitude only,

F = 106.7 N

5 0
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Answer:

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