A 20 g ball is fired horizontally with initial speed v0 toward a 95 g ball that is hanging motionless from a 1.4 m long string. The balls undergo a head-on, perfectly elastic collision, after which the 95 g ball swings out to a maximum angle max = 50°. What was v0?

**Answer 1**

For 95g ball:

since:(1/2)mv^2=mgh & (h=1.4 –1.4*cos(50)= 0.5m)

(1/2)95V^2=95*0.98*0.5

therefore V1=0.99m/s

Since perfectly elastic collision, then Ma*V0+0=0+MbV1

therefore 20*V0=95*0.99

V0=4.702m/s

**Answer 2**

(1/2)m2v2^2 = m2g∆h

v2^2 = 2g∆h

v2 = √(2g∆h)

m1v0 = m2v2 + m1v1

m1v1 = m1v0 – m2v2

v1 = v0 – (m2/m1)v2

v1 = v0 – (m2/m1)√(2g∆h)

(1/2)m1v0^2 = (1/2)m1v1^2 + (1/2)m2v2^2

v0^2 = v1^2 + (m2/m1)g∆h

v0^2 = (v0 – (m2/m1)√(2g∆h))^2 + (m2/m1)g∆h

v0^2 = v0^2 – 2v0(m2/m1)√(2g∆h) + (m2/m1)^2(2g∆h) + (m2/m1)g∆h

2v0(m2/m1)√(2g∆h) = (m2/m1)^2(2g∆h) + (m2/m1)g∆h

2v0√(2g∆h) = (m2/m1)(2g∆h) + g∆h

2v0 = (m2/m1)(2g∆h)/√(2g∆h) + g∆h/√(2g∆h)

2v0 = (m2/m1)√2g∆h) + (1/2)√(2g∆h)

v0 = (1/2)√(2g∆h)((m2/m1) + 1/2)

∆h = 1.4(1 – cos50°)

v0 = (1/2)√(2)(9.80665)(1.4)(1 – cos50°)))((95/20) + 1/2)

v0 ≈ 6.8159 m/s