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sdas [7]
3 years ago
5

Suppose you are watching an ice-hockey game with some classmates. As the puck is sliding across the ice at a relatively constant

speed, it passes close to a player, who hits it (with his stick) in the same direction as it is already moving. After the hit, you notice that the puck is moving much faster than it was before the hit. Your classmates propose the following three explanations for why the puck is moving faster after the hit than before. Whose reasoning do you agree with?
Student A: "The force of the stick was transferred to the puck during the hit. After the hit, the puck has more force, so it moves faster."
Student B: "The puck still has the force that started it moving, which is what keeps it moving at a constant speed. When the player hit it, he adds to this force, which makes the speed of the puck increase. After the hit, this extra force is gone, so the puck stops increasing its speed, but the original force is still there so now it moves at a faster constant speed."
Student C: "While the stick is in contact, it applies a force to the puck that makes its speed increase. As soon is contact is lost, this force disappears, so the speed of the puck stops increasing."
Physics
1 answer:
zavuch27 [327]3 years ago
5 0

Answer:

the answer, the correct one is C

Explanation:

Let's propose the solution of this problem to know which explanation is correct, when the concha stick with the disc is an impulse exercise

                 

             I = ΔP

            ∫ F dt = pf-po

             ∫ F dt = m v_f - m v₀

Therefore, during the time that the contact lasts, a force is applied to the disk, which causes that if the amount of movement increases and therefore its speed increases, when the constant ceases the forces are reduced to zero and the disk no longer changes its momentum following with constant velocity.

When reviewing the answer, the correct one is C

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A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte
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Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

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