A 4.70 kg chunk of ice is sliding at 15.0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 4.70 kg chunk of ice that is initially at rest. Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down.
How fast are they moving when they reach the valley floor again?
By the conservation of momentum
As there is no friction,
So no energy is lost there.
So the speed is the same, 7.5 m/s
P.S. due to inelastic collision, kinetic energy is not conserved here.
Well, well, well... looks like those ice chunks decided to go on a little adventure, huh? I guess they got tired of being cold and wanted to warm up in the valley. Now, let me calculate their speed when they reach the valley floor again.
Since we're dealing with a collision, we can apply the conservation of momentum. The initial momentum of the first ice chunk is given by its mass (4.70 kg) multiplied by its velocity (15.0 m/s). And since the second ice chunk is initially at rest, its initial momentum is zero.
Now, after the collision, the two ice chunks stick together and move as a single unit. Let's call their final speed when they reach the valley floor "vf".
Since momentum is conserved, the total momentum before the collision should be equal to the total momentum after the collision. Mathematically, this can be expressed as:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
Plugging in the values, we get:
(4.70 kg * 15.0 m/s) + (4.70 kg * 0 m/s) = (4.70 kg + 4.70 kg) * vf
70.5 kg*m/s = 9.4 kg * vf
Now, let's solve for vf:
vf = 70.5 kg*m/s / 9.4 kg
vf ≈ 7.5 m/s
So, my funny friend, when those adventurous ice chunks reach the valley floor again, they will be sliding at approximately 7.5 m/s. I hope they enjoy their little roller coaster ride!
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the initial velocity of the sliding ice chunk as v1 and the final velocity (after collision) as v2.
Mass of each ice chunk, m = 4.70 kg
Initial velocity of the sliding ice chunk, u1 = 15.0 m/s
Final velocity (after collision) of both ice chunks, v2 = ?
Applying the conservation of momentum:
Total momentum before the collision = Total momentum after the collision
(m1 * u1) + (m2 * u2) = (m1 * v2) + (m2 * v2)
Here, m1 and m2 are the masses of the sliding chunk and the stationary chunk, respectively. u2 is the initial velocity of the stationary ice chunk, which is 0 since it is at rest.
Plugging in the given values:
(4.70 kg * 15.0 m/s) + (4.70 kg * 0) = (4.70 kg + 4.70 kg) * v2
(70.5 kg·m/s) = (9.40 kg) * v2
v2 = (70.5 kg·m/s) / (9.40 kg)
v2 ≈ 7.5 m/s
So, when the ice chunks reach the valley floor again, they will be moving at approximately 7.5 m/s.
To answer this question, we can use the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant if no external forces act on it.
Let's denote the initial velocity of the first chunk of ice as v1 (15.0 m/s) and the initial velocity of the second chunk of ice as v2 (0 m/s since it is initially at rest). The masses of both chunks of ice are given as 4.70 kg.
The initial momentum of the system is the sum of the individual momenta of the two chunks of ice, which can be calculated using the formula:
Initial momentum = (mass of ice 1 * velocity of ice 1) + (mass of ice 2 * velocity of ice 2)
Initial momentum = (4.70 kg * 15.0 m/s) + (4.70 kg * 0 m/s)
Next, since the two chunks of ice collide and stick together, they move as a single unit with a combined mass of 4.70 kg + 4.70 kg = 9.40 kg.
Let's denote the final velocity of the system as vf.
The final momentum of the system can be calculated as:
Final momentum = (mass of ice 1 + mass of ice 2) * final velocity
Final momentum = 9.40 kg * vf
According to the principle of conservation of momentum, the initial momentum and final momentum should be equal. Therefore, we can equate the initial and final momentum to solve for vf:
(4.70 kg * 15.0 m/s) + (4.70 kg * 0 m/s) = 9.40 kg * vf
Now we can solve this equation to find the final velocity (vf).
inertia before collision:
P = 70*15 + 4.7 * 0
Inertia after collision also =
P = (74.7)v
kinetic energy after collision = Ke = (1/2)(74.4)v^2, do not bother to compute because
at max distance up slope 74.4 g h = that Ke
and when it comes back down
Ke is that Ke again
v = -v