L1 M1 2 M2 A hanmer of mass Mi-2.00 kg on a massless stick of length Li = 30.0 c

Question

L1 M1 2 M2 A hanmer of mass Mi-2.00 kg on a massless stick of length Li = 30.0 cm is nailed to a wall, so that it is free to turn around the axis (se Figure). Starting at rest from the horizontal position, it swings down and hits a ball of mass m = 0.100 kg in an elastic collision. The ball is initially at rest, and when hit proceeds to slide frictionlessly along the table until it reaches a second hannner of mass M,-1.50 kg on a massless stick of length L2 20.0 cm. In this collision, the ball attaches itself to the hammer, and the hammer and ball swing up to a height h a) What is the speed of the ball as it slides along the table? (Expression and number, please!) b) What is the height h? (Expression and number, please!) c) How much of the initial energy in the hammer Mi ends up in the ham mer/ball (M2 m) system at the end (as mechanical energy)? (Expression and number, please!)

Explanation / Answer

initial mechanical energy of M1, E1 = M1*g*L1

final mechanical energy of M1 before hitting ball Ef = (1/2)*m*v1^2

Ef = Ei

(1/2)*M1*v1^2 = M1*g*L1

v1 = sqrt(2*g*L1) = sqrt(2*9.8*0.3) = 2.42 m/s

during elastic collision

momentum before collision = momentum after collision

speed of M1, v1 = 2.42 m/s

speed of m , u1 = 0

IN ELASTIC COLLISION

M1 = 2 kg                    m = 0.1 kg

speeds before collision

v1i = 15 m/s                   v'i = 0 m/s

speeds after collision

v1f = ?                         v'f = ?

initial momentum before collision

Pi = m1*v1i + m*v'i

after collision final momentum

Pf = m1*v1f + m*v'f

from momentum conservation

total momentum is conserved

Pf = Pi

M1*v1i + m*v'i = M1*v1f + m*v'f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*M1*v1i^2 + 0.5*m*v'i^2

KEf =   0.5*M1*v1f^2 + 0.5*m2*v'f^2

KEi = KEf

0.5*M1*v1i^2 + 0.5*m2*v'i^2 = 0.5*M1*v1f^2 + 0.5*m*v'f^2 .....(2)

solving 1&2

we get

v1f = ((M1-m)*v1i + (2*m*v'i))/(M1+m)

v'f = ((m-M1)*v'i + (2*M1*v1i))/(M1+m)

v'f =   ((m-M1)*v'i + (2*M1*sqrt(2*g*L1))/(M1+m)

v'f = (0 + (2*2*sqrt(2*9.8*0.3)))/(2+0.1)

v'f = 4.62 m/s

===========================

the ball undergoes Inelastic collision wit hammer 2

momentum before collision = momentum after collision

m*v'f = (M2+m)*V2

v2 = m*v'f/(M2+m)

v2 = 0.1*4.62/(1.5+0.1) = 0.289 m/s

Kinetic energy after collision = potential energy at height h

(1/2)*(M2+m)*v2^2 = (M2+m)*g*h

h = v2^2/(2*g)

h = 0.289^2/(2*9.8) = 0.0043 m

===================

c)

mechanical energy of M1 = M1*g*L1

mechanical energy of M2+m = (M2+m)*g*h

(M2+m)*g*h/(M1*g*L1) = (1.5+0.1)*0.0043/(2*0.3) = 0.0115

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