Download Problem 16.40 The 1-kg ball is given a horizontal velocity of 1.2 m/s

Problem 16.40 The 1-kg ball is given a horizontal
velocity of 1.2 m/s at A. Photographic measurements
indicate that b = 1.2 m, h = 1.3 m, and the duration of
the bounce at B is 0.1 s. What are the components of
the average impulsive force exerted on the ball by the
floor at B?
y
A
2m
h
Solution: The impulse is
x
B
t2
b
F dt = Fave (t2 − t1 ) = m(v2 − v1 ).
t1
Subtract the weight of the ball from the average impulsive force:
The velocities are determined from the path (from Newton’s second
√
law for free fall): v1 = 1.2i − 2gyj = 1.2i − 6.26j (m/s). The verti√
cal velocity after the bounce at B is 2gh = 5.05 m/s. The time of
flight after the bounce at B is twice the time required to fall a distance
h, from which
Fimp = Fave − (−mgj) = −0.345i + 123j (N)
y
2h
= 1.03 s.
t =2
g
2m
The horizontal velocity after the bounce at B is
h
B
b
/ t = 1.17 m/s,
ovrbt
√
v2 = 1.2i +
x
b
2 ghj = 1.17i + 5.05j (m/s).
From which
t2
F dt = Fave (t2 − t1 ) = −0.0345i + 11.31j N-s.
t1
Fave =
1
(−0.0345i + 11.31j) = −0.345i + 113.1j (N).
0.1
Problem 16.41 At time t = 0, the two masses are
released from rest on the smooth surface with the spring
stretched. Show that at any later time t, the velocities of
the masses are related by
k
mB
mA
mA vA + mB vB = 0.
Strategy: Write the principle of impulse and momentum for each mass.
Solution: The force on mass A is equal and opposite to that on
mass B. Hence, we can write
FA + FB = 0
and
t
FA dt +
0
0
t
FB dt =
t
0 dt = 0.
0
From the principle of impulse and momentum, we can write
t
v
FA dt = mA
0
dv = mA vA .
0
Similarly, we can write
0
t
FB dt = mB
v
k
mA
dv = mB vB .
0
Substituting into the equation above, we get mA vA + mB vB = 0, as
was required.
mB