Two blocks of masses m1 = 2.00 kg and m2-4.35 kg are released from rest at a hei

Question

Two blocks of masses m1 = 2.00 kg and m2-4.35 kg are released from rest at a height of h 5.40 m on a frictionless track as shown in the figure below. When they meet on the level portion of the track, they undergo a head-on, elastic colllslon. Determine the maxlmum helghts to which m and m2 rise on the curved portlon of the track after the collislon hml= hm2= 18.7 Our response differs from the correct answer by more than 10%. Double check your calculations, m 0.317 Your response differs from the correct answer by more than 10%. Double check your calculations, m 1

Explanation / Answer

By the law of energy conservation

KE(final) = PE(initial)

1/2mu^2 = mgh

u = sqrt(2gh)

u1 and u2 = sqrt[2 * 9.8 * 5.4] = 10.28 m/s

Now by the law of momentum conservation

Let we take the direction of m1 as +ve

m1u1 + m2u2 = m1v1 + m2v2

2 * 10.28 + 4.35 * (-10.28) = 2v1 + 4.35v2

2v1 + 4.35v2 = -24.16 eq1

For elastic collision,

v1 - v2 = u2 - u1

v1 - v2 = -10.28 - 10.28

v1 - v2 = -20.56 eq2

By eq1 + 4.35 * eq2,

6.35v1 = -24.16 + (4.35* (-20.56))

v1 = -113.596/6.35

= -18 m/s

v2 = v1 + 20.56

= -18 +20.56 = 2.56 m/s

Again by the law of energy conservation,

PE(final) = KE(initial)

mgh = 1/2mv^2

h = v^2/2g

h1 = v1^2/2g = 18^2/(2*9.8) = 16.53 m

h2 = v2^2/2g = 2.56^2/(2*9.8) = 0.33 m

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