July 11, 2008
PHY2053 Discussion
Quiz 6 (Chapter 6)
Name:
UFID:
*1. (5pts) A 0.800-kg steel ball strikes a massive wall at 15.0 m/s at an angle of 60.0˚
with the plane of the wall. It bounces off the wall with the same speed and angle. If the
ball is in contact with the wall for 0.0500 s, what is the average force exerted by the wall
on the ball?
Taking the +x direction perpendicular to and into the wall and +y along the plane of the
wall, the initial and final velocity of the ball are
vi = (15cos30˚, 15sin30˚) = (13.0 m/s, 7.5 m/s)
vf = (-15cos30˚, 15sin30˚) = (-13.0 m/s, 7.5 m/s)
Thus the change in momentum is given by
Δp = m(vf-vi) = 0.8×(-13-13, 7.5-7.5) = (-20.8 kgm/s, 0 kgm/s)
According to the impulse-momentum theorem, this equals impulse I = FΔt, thus
F = I/Δt = (-20.8, 0)/0.05 = (-416 N, 0N)
The average force is 416 N perpendicular to and out of the wall.
***2. (5pts) A 50.0 kg-person running on the ground jumps onto a 150-kg plank, which is
initially at rest on a frictionless frozen pond. The person slides on the plank with an
initial speed of 4.00 m/s and finally comes to rest relative to the plank. The coefficient of
kinetic friction between the person and the plank is 0.200. How long does it take the
person to stop relative to the plank?
Since the system consists of the person and plank is isolated, the total momentum is
conserved. The person comes to rest relative to the plank in the end, which means the
final velocities of the plank and the person are equal. We have
mvi = (m+M)vf ⇒ vf = mvi/(m+M) = 50×4/(50+150) = 1.00 m/s
So the velocity of the person changes by Δv = vf-vi = 1-4 = -3.00 m/s. Applying the
Newton’s 2nd law to the person, the acceleration of the person is
ma = -μmg a = -μg = -1.96 m/s²
Thus the time interval is given by
Δv = aΔt ⇒ Δt = Δv/a = -3/-1.96 = 1.53 s
*3. (5pts) A 10.0-g bullet is fired into a ballistic pendulum, which consists of a 1.00-kg
wooden block suspended from light wires. The bullet is stopped by the block, and the
entire system swings up to a height of 15.0 cm. Find the initial speed of the bullet.
After the collision, the mechanical energy of the system is conserved. Thus the velocity
just after the collision is given by
(1/2)mvf² = (M+m)gh ⇒ vf = √(2gh) = √(2×9.8×0.15) = 1.71 m/s
Since the total momentum is conserved during the collision, we have
mvi = (m+M)vf ⇒ vi = (m+M)vf/m = 1.01×1.71/0.01 = 173 m/s
**4. (5pts) A 2.00-kg block slides on a frictionless table of height 1.00 m and collides
elastically with a 1.00-kg block initially at rest at the edge of the table. After the impact
the 1-kg block lands 2.00 m from the bottom of the table. How far away from the bottom
of the table does the 2-kg block land?
The initial velocity of the projectile motion of the 1-kg block, which is equal to the
velocity just after the collision, is given by
Δy = -(1/2)gt² ⇒ t = √(-2Δy/g) = √(-2×-1/9.8)= 0.452 s
Δx2 = v2ft ⇒ v2f =Δx2/t = 2/0.452 = 4.42 m/s
Since the collision is elastic, both the momentum and relative speed is conserved. Thus
the velocity of the 1-kg block just after the collision is given by
m1v1i = m1v1f+m2v2f & v1i = v2f-v1f ⇒ m1(v2f-v1f) = m1v1f+m2v2f
⇒ v1f = (m1-m2)v2f/2m1 = (2-1)4.42/(2×2) = 1.105 m/s
The flight time of the 2-kg block is the same as that of the 1-kg block, thus
Δx1 = v1ft = 1.105×0.452 = 0.500 m