Q1 a ball of mass, m, swings through a height, h, of 0.03m toward a ball of mass

Question

Q1

a ball of mass, m, swings through a height, h, of 0.03m toward a ball of mass, M =3m, which is initially at rest. If m has an  initial speed, v, of 5 m/s and strikes M elastically, determine how high each ball rises afterward neglecting air resistance?

Q2:

A bullet emerges  from the muzzle of agun with a speed of 300m/s. The resultant force on the bullet, while it is in the barrel, is given by: F=400 N-1/3(4*10^5 N/s)t, where t is in seconds

a) calculate the time for the bullet to travel the lenght of the barrel, assuming the force is zero at the end of the barrel?

b) find the impulse of the force ?

c)find the mass of the bullet?

Explanation / Answer

1. Initial speed of m when it strikes M = (5^2 + 2*9.8*0.03)^0.5 = 5.06 m/s

Conservation of momentum gives

Mv2 + mv1 = mu1

3v2 + v1 = 5.06 ---1

From coefficient of restitution = 1

v2 - v1/ u1 - u2 = 1

v2 - v1 = 5.06 ----2

Adding 1 and 2

4v2 = 10.12

v2 = 2.53 m/s

v1 = -2.53 m/s

Heights they rise to = h = v^2/2g = 0.33 m each in opposite directions

2. a. F = 0

t = 3*400/4x10^5 = 0.003 s

b. Impulse, J = Integral Fdt = 400t - 1/3(2x10^5)t^2 between t = 0 and t = 0.003

J = 0.6 kgm/s

c. Mass of the bullet = 0.6/300 = 2 grams

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