Question

A 25. As shown in figure 9, a ball of mass M is hanging from a rope to make a pendulum. A 10 g bullet strikes the ball with a speed v = 308 m/s. The center of mass of the ball + bullet rises a vertical distance of h = 12 cm. Assuming that the bullet remains embedded, calculate the mass M of the ball. (Ans: 2.0 kg)

Answer 1

25.

using energy conservation between initial and final position of pendulum after collision

KEi + PEi = KEf + PEf

PEi = 0,

KEf = 0, at highest point velocity will be zero.

KEi = PEf

0.5*m1*V1^2 = m1*g*h

h = vertical displacement = 12 cm = 0.12 m

V1 = final speed of ball+bullet just after collision = sqrt (2*g*h)

V1 = sqrt (2*9.81*0.12) = 1.5344 m/sec

Now using momentum conservation before and after collision

Since this completely inelastic collision, So

Pi = Pf

m1*u1 + m2*u2 = (m1 + m2)*V1

u1 = initial speed of ball = 0 m/sec

u2 = Initial speed of bullet = 308 m/s

m2 = mass of bullet = 10 gm = 0.01 kg

m1 = mass of ball = M

So,

M*0 + 0.01*308 = (M + 0.01)*1.5344

M = (0.01*308/1.5344) - 0.01 = 1.9973 = 2.0 kg

M = Mass of ball = 2.0 kg

27.

Using momentum conservation before and after collision:

Pi = Pf

m1*v1 + m2*v2 = (m1 + m2)*V

V = (m1*v1 + m2*v2)/(m1 + m2)

V = (2.0*50 + 5*(-20))/(2 + 5) = 0/7

V = 0 m/s

final speed of both mass after collision is zero, which means all the kinetic energy is lost.

KE_lost = KEi

KE_lost = (1/2)*m1*v1^2 + (1/2)*m2*v2^2

KE_lost = (1/2)*2.0*50^2 + (1/2)*5.0*(-20)^2

KE_lost = 3500 J

Let me know if you've any query.