- #1

Soaring Crane

- 469

- 0

A.4490 N----b. 3360 N------------c. 7850 N-----d. 6720 N------e. 5590 N

If t_total = 7.00 s and time of collision is 0.5 ms*(1 s/1000 ms) = 5*10^-4 s, then 7.00 s – 5*10^-4 s = 6.9995 s, which is the time that is not dedicated to the contact time.

Since the ball rebounds, it takes 6.9995 s/2 = 3.49975 s for the ball to travel from rest to the plate another 3.49975 s for the ball to travel from the plate back to the original height.

F = m*(delta v)/t_contact

To find the ball’s velocity, I though of the conservation of energy,

Where v = sqrt(2*g*h).

I also thought of the kinematic equation v_f = v_o + a*t in which a = 9.8 m/s^2 if the ball is dropped.

Now, would v = 9.8 m/s^2*(3.49975 s) = 34.29755 m/s? (same value if h = 0.5a*t^2 used)

F = (.049 kg)*(34.29755 m/s)/(5*10^-4 s) = 3361.2 J ?

A 18 g bullet is shot vertically into an 10 kg block. The block lifts upward 9 mm. The bullet penetrates the block in a time interval of 0.001 s. Assume the force on the bullet is constant during penetration. The initial kinetic energy of the bullet is closest to:

a.490 J

b.0.88 J

c.250 J

d.330 J

e.0.0016 J

First off, I was puzzled at why a time interval was given. “Does it possibly have something to do with the impulse?” I thought.

Here is what I ended up doing. Please check to see if the setup is correct.

Inelastic collision: m_bullet*v = (m_bullet + m_block)*v’

For block: For conservation of energy, mgh = 0.5*m*v^2 ??

v = sqrt(2*g*h) = sqrt(2*9.8*0.009 m) = 0.42 m/s

So v_block = v’?

V_bullet = [(m_bullet + m_block)*v’]/[ m_bullet] = [(0.018 kg + 10 kg)*0.42 m/s]/[0.018 kg] = 233.753 m/s

KE = (mv^2)/2 = 0.5*(0.018 kg)*(233.753 m/s)^2 = 492 J (choice a.)

Thanks.