A small block (50g) slides across a frictionless table with a speed of 2.85 m/s.

Question

A small block (50g) slides across a frictionless table with a speed of 2.85 m/s. The block collides with a clay pendulum bob, which is initially hanging from a vertical string. The block sticks to the pendulum bob and they begin to swing together, moving an initial speed after the collision of .678 m/s. What is the mass of the pendulum bob? It takes .45 seconds for the pendulum to return (for the first time) to its vertical position after the collision. How long is the pendulum's string? What is the maximum angle from the vertical that the pendulum reaches in its motion? When the pendulum bob gets back to the lowest point of its motion, how fast is it moving? (Don't overthink it!) When it gets to that point, what is the tension in the string?

Explanation / Answer

given

m = 50 g = 0.05 kg

v = 2.85 m/s

V = 0.678 m/s

M = ?

a) Apply conservation of momentum

m*v = (m + M)*V

==> M = m*v/V - m

= 0.05*2.85/0.678 - 0.05

= 0.160 kg or 160 grams

b) so, Time period, T = 2*0.45

= 0.9 s

let L is the length of the pendulum,

we know, T = 2*pi*sqrt(L/g)

==> L = g*T^2/(4*pi^2)

= 9.8*0.9^2/(4*pi^2)

= 0.201 m

c) h = V^2/(2*g)

= 0.678^2/(2*9.8)

= 0.02345 m

let theta is the maximum angle made by string with vertical.

use,

h = L*(1 - cos(theta))

h/L = 1 - cos(theta)

cos(theta) = 1 - h/L

cos(theta) = 1 - 0.02345/0.201

cos(theta) = 0.883

theta = cos^-1(0.883)

= 28 degrees

d) same. i.e 0.678 m/s

e) T = (m+M)*(g + V^2/L)

= (0.05 + 0.160)*(9.8 + 0.668^2/0.201)

= 2.52 N

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