Question

1) A 0.490 kg soccer ball is kicked and hits a 2.45 kg stationary tether ball hanging from a 1.49 m tether resulting in a perfectly elastic collision. If the tether ball swings out at a maximum angle of 45.4°, what was the soccer ball's velocity before the collision?
2) A 5.27 g pellet is shot horizontally from a BB gun at a speed of 15.2 m/s into a 21.1 g wooden block. The wooden block is attached to a spring and lies on a frictionless table. If the collision is inelastic and the spring constant k = 20.0 N/m, what is the maximum compression of the spring?

Answer 1

1)

let m1 = 0.490 kg
m2 = 2.45 kg
L = 1.49 m
theta = 45.4 degrees
vertical height raised by the tether ball,
h = L*(1 - cos(theta))
= 1.49*(1 - cos(45.4))

= 0.4438 m

velocity of m2 after the collsion,
v2 = sqrt(2*g*h)

= sqrt(2*9.8*0.4438)

= 2.95 m/s

let u1 is the velocity of soccer ball before the collision.

we know, v2 = 2*m1*u1/(m1 + m2)

==> u1 = v2*(m1 + m2)/(2*m1)

= 2.95*(0.49 + 2.45)/(2*0.49)

= 8.85 m/s <<<<<<<<<<-------------------Answer

2)

let m1 = 5.27 g = 0.00527 kg
u1 = 15.2 m/s
m2 = 21.1 g = 0.0211 kg

let v is the speed of the wooded block just after the collision.

Apply conservation of momentum

final momentum = initial momentum

(m1 + m2)*v = m1*u1

v = m1*u1/(m1 + m2)

= 5.27*15.2/(5.27 + 21.1)

= 3.04 m/s

let x is the maximum compression.

now Apply conservation of energy

(1/2)*k*x^2 = (1/2)*(m1 + m2)*v^2

x = v*sqrt((m1 + m2)/k)

= 3.04*sqrt((0.00527 + 0.0211)/20 )

= 0.110 m <<<<<<<<-------------------Answer