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Direct impact of two spheres

Short answer

Applying momentum conservation and restitution e = 0.7 yields v_A = 3.45 m/s, v_B = 6.25 m/s. Kinetic energy loss is 7.65 J. Impulse on B equals its momentum change: 3×(6.25 –2) = 12.75 N·s.

A sphere A of mass 5 kg moving at 6 m/s overtakes a sphere B of mass 3 kg moving at 2 m/s in the same straight line. The coefficient of restitution is 0.7. Find (a) the velocities of both spheres after impact, (b) the loss of kinetic energy, and (c) the impulse exerted on B.

Given: mA=5m_A = 5 kg, uA=6u_A = 6 m/s, mB=3m_B = 3 kg, uB=2u_B = 2 m/s, e=0.7e = 0.7 (both to the right).

(a) Velocities after impact

Conservation of momentum: total momentum =5(6)+3(2)=36.0= 5(6) + 3(2) = 36.0 kg·m/s.

5vA+3vB=36(1)5v_A + 3v_B = 36 \quad (1)

Restitution: vB−vA=e(uA−uB)=0.7(6−2)=2.8v_B - v_A = e(u_A - u_B) = 0.7(6 - 2) = 2.8 (2)

From (2), vB=vA+2.8v_B = v_A + 2.8. Put in (1): 5vA+3vA+8.4=365v_A + 3v_A + 8.4 = 36, so 8vA=27.68v_A = 27.6.

vA=3.45 m/svB=6.25 m/sv_A = 3.45\ \text{m/s} \qquad v_B = 6.25\ \text{m/s}

Both move to the right; B now moves faster than A, so there is no further impact.

(b) Loss of kinetic energy

Before (J)After (J)
A12(5)(6)2=90\tfrac12(5)(6)^2 = 9012(5)(3.45)2=29.76\tfrac12(5)(3.45)^2 = 29.76
B12(3)(2)2=6\tfrac12(3)(2)^2 = 612(3)(6.25)2=58.59\tfrac12(3)(6.25)^2 = 58.59
Total96.0088.35
ΔT=96.00−88.35=7.65 J\Delta T = 96.00 - 88.35 = 7.65\ \text{J}

Check by formula: 5×32×8(1−0.49)(4)2=0.9375×0.51×16=7.65\dfrac{5 \times 3}{2 \times 8}(1 - 0.49)(4)^2 = 0.9375 \times 0.51 \times 16 = 7.65 J.

(c) Impulse on B

Impulse equals change of momentum of B:

I=mB(vB−uB)=3(6.25−2)=12.75 N⋅sI = m_B(v_B - u_B) = 3(6.25 - 2) = 12.75\ \text{N·s}

(The impulse on A is equal and opposite, 5(3.45−6)=−12.755(3.45 - 6) = -12.75 N·s.)

Answer: v_A = 3.45 m/s, v_B = 6.25 m/s; energy lost = 7.65 J; impulse on B = 12.75 N·s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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