Momentum and collision calculator (elastic and inelastic)

Momentum and collision calculator: final velocities, total momentum and kinetic energy lost in an elastic, inelastic or partly elastic head-on collision.

Cập nhật Ví dụ đã kiểm tra: 5

Signed: positive is to the right
Body 2 starts to the right of body 1
Thử
Velocity of body 1 after
m/s
Velocity of body 1 after: 0.333333 m/s
Số chữ số có nghĩa: 6; Gần nhất; nếu cách đều thì ra xa số không
Velocity of body 2 after
4.33333m/s
Total momentum (conserved)
5kg·m/s
Kinetic energy before
9.5J
Kinetic energy after
9.5J
Kinetic energy lost
0J
Share of kinetic energy lost
0.00%
Impulse on body 2
5.33333N·s

In this elastic collision momentum stays at 5 kg·m/s; body 1 leaves at 0.333333 m/s and body 2 at 4.33333 m/s, and 0% of the kinetic energy is lost.

Velocity before and after (m/s)

Body 1 before: 33Body 1 beforeBody 1 after: 0.33330.3333Body 1 afterBody 2 before: −1−1Body 2 beforeBody 2 after: 4.3334.333Body 2 after

Kinetic energy (J)

0246810BeforeAfter
Body 1Body 2Lost to heat, sound, deformation
Cách tính S
  1. Momentum before (conserved)

    p=m1u1+m2u2=(2)(3)+(1)(−1)=5 kg⋅m/sp = m_1u_1 + m_2u_2 = (2)(3) + (1)(-1) = 5\ \mathrm{kg\cdot m/s}
  2. Restitution

    e=v2−v1u1−u2=1e = \frac{v_2 - v_1}{u_1 - u_2} = 1

    e = 1 keeps all kinetic energy; e = 0 means the bodies move off together.

  3. Final velocities

    v1=p+m2e(u2−u1)m1+m2=0.333333 m/s,v2=p+m1e(u1−u2)m1+m2=4.33333 m/sv_1 = \frac{p + m_2e(u_2 - u_1)}{m_1 + m_2} = 0.333333\ \mathrm{m/s},\quad v_2 = \frac{p + m_1e(u_1 - u_2)}{m_1 + m_2} = 4.33333\ \mathrm{m/s}
  4. Kinetic energy

    KEbefore=9.5 J,KEafter=9.5 J,ΔKE=0 JKE_{\text{before}} = 9.5\ \mathrm{J},\quad KE_{\text{after}} = 9.5\ \mathrm{J},\quad \Delta KE = 0\ \mathrm{J}

Giới thiệu Momentum and collision calculator (elastic and inelastic)

In a collision between two bodies, the total momentum m₁u₁ + m₂u₂ is the same before and after. The coefficient of restitution e, the ratio of separation speed to approach speed, supplies the second equation: e = 1 is a perfectly elastic collision and e = 0 means the bodies stick together. The two equations give both final velocities, and comparing ½mv² before and after gives the kinetic energy lost.

The default, a 2 kg body at 3 m/s meeting a 1 kg body moving at −1 m/s elastically, sends them off at 0.333 m/s and 4.333 m/s. A 1,000 kg car at 20 m/s that locks onto a parked 1,500 kg car moves off at 8 m/s, and 60% of the kinetic energy goes into deformation, heat and sound.

Motion is along one line: choose a positive direction and give velocities the other way a minus sign. External forces such as road friction are taken as negligible during the impact.

Ví dụ có lời giải

Equal masses, elastic: velocities swap

Collision
Elastic
Mass of body 1
1 kg
Velocity of body 1 before
2 m/s
Mass of body 2
1 kg
Velocity of body 2 before
0 m/s
Show velocities in
m/s
Velocity of body 1 after
0 m/s
Velocity of body 2 after
2 m/s
Kinetic energy lost
0 J

Nguồn đối chiếu: OpenStax UP1 §9.4: equal-mass elastic collision exchanges velocities

Car hits a parked car and they lock

Collision
Perfectly inelastic
Mass of body 1
1000 kg
Velocity of body 1 before
20 m/s
Mass of body 2
1500 kg
Velocity of body 2 before
0 m/s
Show velocities in
m/s
Velocity of body 1 after
8 m/s
Velocity of body 2 after
8 m/s
Kinetic energy before
200,000 J
Kinetic energy after
80,000 J
Share of kinetic energy lost
60.00%

Nguồn đối chiếu: Python 3.8 decimal: v = 20000/2500 = 8; KE 200000 → 80000 J

e = 0.5 head-on

Collision
Coefficient e
Coefficient of restitution e
0.5
Mass of body 1
2 kg
Velocity of body 1 before
3 m/s
Mass of body 2
1 kg
Velocity of body 2 before
-1 m/s
Show velocities in
m/s
Velocity of body 1 after
1 m/s
Velocity of body 2 after
3 m/s
Total momentum (conserved)
5 kg·m/s
Kinetic energy lost
4 J

Nguồn đối chiếu: Python 3.8 fractions: v1 = (5 + 1·0.5·(−4))/3 = 1, v2 = (5 + 2·0.5·4)/3 = 3; KE 9.5 → 5.5 J

Heavy ball hits a light one, elastic

Collision
Elastic
Mass of body 1
10 kg
Velocity of body 1 before
1 m/s
Mass of body 2
1 kg
Velocity of body 2 before
0 m/s
Show velocities in
m/s
Velocity of body 1 after
0.818182 m/s
Velocity of body 2 after
1.81818 m/s

Nguồn đối chiếu: Python 3.8 fractions: v1 = 9/11, v2 = 20/11

Câu hỏi

What is the difference between elastic and inelastic collisions?

Both conserve momentum; only an elastic collision also conserves kinetic energy. In a perfectly inelastic collision the bodies stick together and lose the most kinetic energy that momentum conservation allows. When a 1,000 kg car at 20 m/s locks onto a parked 1,500 kg car, momentum stays at 20,000 kg·m/s while kinetic energy falls from 200 kJ to 80 kJ. Most real collisions fall between the two extremes.

What is the coefficient of restitution?

It is the relative speed after a collision divided by the relative speed before, e = (v₂ − v₁) ÷ (u₁ − u₂), a number from 0 to 1. For a ball dropped onto a rigid floor, e = √(bounce height ÷ drop height). The ITF requires a type 2 tennis ball dropped from 254 cm onto concrete to rebound 135–147 cm, which corresponds to e between 0.73 and 0.76.

How do you calculate momentum?

Momentum is mass times velocity, p = mv, measured in kg·m/s. It has a direction, so velocities in opposite directions carry opposite signs. A 1,000 kg car at 20 m/s has 20,000 kg·m/s. With no outside forces the total is the same before and after a collision, which is why two 2 kg carts meeting head-on at 5 m/s and sticking together stop dead.

What happens when two equal masses collide elastically?

They swap velocities. A 1 kg ball at 2 m/s hitting an identical ball at rest stops, and the second ball leaves at 2 m/s with all the kinetic energy; a Newton's cradle shows the same effect. With unequal masses the lighter body leaves faster: a 10 kg ball at 1 m/s sends a 1 kg ball off at 1.82 m/s and slows to 0.82 m/s.

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Độ chính xác phụ thuộc vào dữ liệu nhập và giả định của phương pháp. Phép tính thập phân dùng 50 chữ số có nghĩa, nhưng ước lượng, phương pháp số và dữ liệu nguồn có thể kém chính xác hơn; làm tròn khi hiển thị không loại bỏ các giới hạn đó. Ví dụ có lời giải đã đối chiếu với nguồn độc lập: 5. Ví dụ, “Equal masses, elastic: velocities swap” được kiểm tra bằng OpenStax UP1 §9.4: equal-mass elastic collision exchanges velocities.

Phương pháp này lấy từ đâu?

OpenStax University Physics Volume 1, §9.4 Types of collisions; HyperPhysics — Elastic and inelastic collisions; coefficient of restitution.

Về công cụ tính này

v1=m1u1+m2u2+m2e(u2−u1)m1+m2,v2=m1u1+m2u2+m1e(u1−u2)m1+m2v_1 = \frac{m_1u_1 + m_2u_2 + m_2e(u_2 - u_1)}{m_1 + m_2},\quad v_2 = \frac{m_1u_1 + m_2u_2 + m_1e(u_1 - u_2)}{m_1 + m_2}

Nguồn

  1. OpenStax University Physics Volume 1, §9.4 Types of collisions
  2. HyperPhysics — Elastic and inelastic collisions; coefficient of restitution

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