rvanveshana.com
t = 0.00s
3kg 5kg
Velocities Before / After Block 1: 6 → -2.50 m/s
Block 2 Speed Shift Block 2: -2 → 3.10 m/s

Live Telemetry & Summary

Observe the live values changing in real-time as the simulation runs. Tweak parameters and press **Simulate** to see new calculations.

Time:0.00 s
Block 1 Speed:6.0 m/s
Block 2 Speed:-2.0 m/s
Total Momentum:8.0 kg·m/s
Total KE:64.0 J

Variable Adjuster

Block 1 Mass (m₁)3kg
110
Block 1 Velocity (u₁)6m/s
-1010
Block 2 Mass (m₂)5kg
110
Block 2 Velocity (u₂)-2m/s
-1010
Elasticity (e)0.7
01

Collision Lab

COLL

Two-body collisions conserve linear momentum. The coefficient of restitution determines whether kinetic energy is also conserved (elastic) or partially lost (inelastic).

m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Whiteboard Solver Steps

Step 1

Conservation of Linear Momentum

In the absence of external net forces, the total momentum of the two-body system is conserved before and after collision.

Ptotal=m1u1+m2u2=8.0 kg⋅m/sP_{total} = m_1 u_1 + m_2 u_2 = 8.0 \text{ kg} \cdot \text{m/s}
Step 2

Coefficient of Restitution (Elasticity)

Restitution measures the elasticity of the collision: - e=1.00e = 1.00: Perfectly Elastic (kinetic energy conserved). - e=0.00e = 0.00: Perfectly Inelastic (objects stick together, maximum kinetic energy loss).

e=v2−v1u1−u2=0.70e = \frac{v_2 - v_1}{u_1 - u_2} = 0.70
Step 3

Post-Collision Velocities

Calculated velocities of Block 1 and Block 2 immediately after the collision event.

v1=−2.50 m/s,v2=3.10 m/sv_1 = -2.50 \text{ m/s}, \quad v_2 = 3.10 \text{ m/s}