Question

A 21 kg wood ball hangs from a 2.0-m-long wire. The maximum tension the wire can withstand without breaking is 367 N. A 1.2 kg projectile traveling horizontally hits and embeds itself in the wood ball. What is the largest speed this projectile can have without causing the cable to break?

A 31 g dart is shot straight up at 9.6 m/s. At the same instant, a 20 g ball of cork is dropped from 2.0 m above the dart. What are the speed and direction of the cork ball immediately after it is hit by the dart? Assume the collision is exactly head-on and the dart sticks in the cork.

Answer 1

1) The velocity of the ball and projectile after collision is found from conservation of momentum:

(m1 + m2)*vf = m2*v0

vf = v0*m2/(m1 + m2)

The force on the wire is the sum of the weight of the ball plus the centrifugal force due to the velocity:

T = m1*g + (m1 + m2)*vf²/R

substitute for vf

T = m1*g + (m1 + m2)*v0²*m2²/[(m1 + m2)²*R]

T = m1*g + v0²*m2²*/[(m1 + m2)*R]

v0² = (T - m1*g)*R*(m1 + m2)/m2²

v0 = √[(T - m1*g)*R*(m1 + m2)]/m2

T = 367 N
m1 = 21.0 kg
m2 = 1.2 kg
R = 2.0 m

v0 = 77.23 m/s

2) Let up be defined as positive and down as negative.

The 31 gram dart starts a velocity if 9.6 m/s.
The 20 g ball starts at a velocity of 0 m/s.

At the point of collision the distance covered by the dart is, t is the time between the drop and the collision:

9.6 m/s*t - 4.9 m/s/s * t^2

whereas the distance covered by the cork is:

4.9 m/s/s * t^2

Thus the total distance covered is:

9.6 m/s*t - 4.9 m/s/s * t^2 + 4.9 m/s/s * t^2 = 2.0 m

t = (2.0 m)/(9.6 m/s) =0.208 seconds

At the point of collision the velocity of the dart is:

9.6 m/s - 9.8 m/s/s*(0.208 s) =7.56 m/s

the velocity of the cork at that point is:

-9.8 m/s/s*(0.208 s) = -2.04 m/s

Since the object stuck together:

M(1)*V(1) + M(2)*V(2) = [M(1) + M(2)]*V(f)

That is, the momentum is conserved.

0.031 kg * 7.56 m/s - 0.020 kg * 2.04 m/s =0.051 kg * V(f)

V(f) = 3.8 m/s