38
kg·m/s
38
kg·m/s
4.75
m/s
4.75
m/s
14
m/s
0
m/s
274
J
90.25
J
-183.75
J
67.06
%
4.75
m/s
-26.25
N·s
26.25
N·s
38
kg·m/s
38
kg·m/s
4.75
m/s
4.75
m/s
14
m/s
0
m/s
274
J
90.25
J
-183.75
J
67.06
%
4.75
m/s
-26.25
N·s
26.25
N·s
The Conservation of Momentum Calculator solves one-dimensional collision problems using the principle that total momentum is always conserved in an isolated system. It handles both perfectly elastic and perfectly inelastic collisions, computing final velocities, kinetic energy changes, and energy loss.
Conservation of momentum states: $$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$$
For a perfectly inelastic collision (objects stick together): $$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$
For a perfectly elastic collision (kinetic energy also conserved): $$v_1' = \frac{(m_1 - m_2)v_1 + 2m_2 v_2}{m_1 + m_2}$$ $$v_2' = \frac{(m_2 - m_1)v_2 + 2m_1 v_1}{m_1 + m_2}$$
This calculator is essential for physics students studying collisions, engineers designing crash safety systems, game developers implementing realistic physics, and anyone working with impact mechanics. It reveals how much kinetic energy survives the collision and what percentage is lost to deformation, heat, and sound.
The calculator applies two conservation laws depending on the collision type:
Perfectly Inelastic: Only momentum is conserved. The objects merge into one body of mass $$m_1 + m_2$$ moving at $$v_f = (m_1 v_1 + m_2 v_2)/(m_1 + m_2)$$. Maximum kinetic energy is lost (converted to heat, deformation, sound).
Perfectly Elastic: Both momentum and kinetic energy are conserved. Solving the two simultaneous equations gives unique final velocities. Equal masses exchange velocities completely: if $$m_1 = m_2$$, then $$v_1' = v_2$$ and $$v_2' = v_1$$.
The kinetic energy before and after is computed as $$KE = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$$, and the energy lost is the difference. In elastic collisions, this difference is zero. In inelastic collisions, the lost energy is $$\Delta KE = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(v_1 - v_2)^2$$.
The total momentum is the same before and after — this is the conserved quantity. For inelastic collisions, both objects share the same final velocity; the percentage of energy lost depends on the mass ratio and relative speed. When a heavy object hits a light stationary one inelastically, little energy is lost; when equal masses collide head-on, up to 100% of kinetic energy can be lost (in the center-of-mass frame). For elastic collisions, check that $$v_1'$$ and $$v_2'$$ swap as expected for equal masses, or that the lighter object bounces back faster than it arrived.
Inputs
Results
A 5 kg object at 10 m/s hits a 3 kg object at -4 m/s. They stick together moving at 4.75 m/s, losing 67% of kinetic energy.
Inputs
Results
Equal masses in elastic collision: the moving object stops and transfers all its velocity to the stationary one — Newton's cradle effect.
Conservation of momentum states that the total momentum of an isolated system (no external forces) remains constant: $$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$$. This holds for all types of collisions — elastic, inelastic, and everything in between. It is one of the most fundamental laws in physics.
In an elastic collision, both momentum and kinetic energy are conserved — no energy is lost. In an inelastic collision, momentum is conserved but some kinetic energy is converted to other forms (heat, sound, deformation). A perfectly inelastic collision is one where the objects stick together, losing the maximum possible kinetic energy.
Only if the total momentum is zero (equal masses moving at equal speeds in opposite directions). In a perfectly inelastic collision, they stick together with zero velocity, converting all kinetic energy to other forms. If total momentum is nonzero, some kinetic energy must remain to carry that momentum.
Newton's cradle demonstrates elastic collisions between equal masses. When one ball strikes the row, it stops and transfers its momentum and energy to the last ball. The intermediate balls barely move because the collision wave propagates through them via elastic deformation. The result is the characteristic one-in-one-out motion.
Yes. In 2D and 3D, momentum is conserved separately in each direction: $$\Sigma p_x = \text{const}$$, $$\Sigma p_y = \text{const}$$. This calculator handles 1D collisions. For 2D oblique collisions, you need to resolve velocities into components and apply conservation in each direction independently.
Explosions and fragmentations also conserve momentum. If a stationary object breaks into pieces, the total momentum of all fragments must be zero (the initial momentum). Each fragment carries momentum, but the vector sum of all momenta remains unchanged. Energy is added from the explosion, so kinetic energy increases.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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