【CH09】Center of Mass and Momentum Homework of ... - 物理學系

Fundamentals of Physics
Halliday & Resnic
東海大學物理系
【Problem 9-33】
Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator
cab free-falls from a height of 36 m. During the collision at the bottom of the elevator shaft, a 90 kg
passenger is stopped in 5.0 ms. (Assume that neither the passenger nor the cab rebounds.) What are
the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the
passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab
hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force
(assuming the same stopping time)?
在電梯撞到前跳起來。在電纜線和安全系統故障後,電梯從 36m 高的地方自由落體落下,
在撞到電梯井底部之前,一位 90 公斤的乘客停在 5ms。問乘客碰撞的(a)衝量(b)平均
施力大小為何?如果在電梯碰到電梯井底部時,乘客以向上速度 7m/s 跳起來碰到電梯天花
板,問(c)衝量(d)平均力大小為何?
1 2 2
:(a) mv = mgh ⇒ v = 2gh = 2(9.8 m/s )(36 m) = 26.6 m/s.
2
3
J = | Δ p| = m| Δ v| = mv=
(90kg)(26.6m/s) ≈ 2.39× 10 N⋅ s.
−3
(b) With duration of Δ t = 5.0× 10 s for the collision, the average force is
3
J 2.39× 10 N ⋅s
avg −3
5
F = = ≈ 4.78× 10 N.
Δ t 5.0× 10 s
(c) If the passenger were to jump upward with a speed of v′ = 7.0 m/s , then the resulting
downward velocity would be
v′′ = v− v′
= 26.6 m/s − 7.0 m/s = 19.6 m/s,
and the magnitude of the impulse becomes
3
J′′ = | Δ p′′ | = m| Δ v′′ | = mv′′
= (90kg)(19.6m/s) ≈ 1.76× 10 N⋅ s.
(d)
3
J′′
1.76× 10 N ⋅s
F′′
avg = = ≈ ×
−3
Δ t 5.0× 10 s
5
3.52 10 N.
【Problem 9-39】
A 91 kg man lying on a surface of negligible friction shoves a 68 g stone away from himself, giving
it a speed of 4 m/s. What speed does the man acquire as a result?
一 91 公斤男子倒臥在一個無摩擦的表面,猛推一個 68 公克的石頭遠離自己,給石頭的速
度為 4m/s,結果這個人獲得怎樣的速度?
: mv m m= mv s s
ms s s (0.068)(4)
−3
vm= = = 3× 10 m/ s
m 91
m
【Problem 9-43】
Figure 9-58 shows a two-ended “rocket” that is initially stationary on a frictionless floor, with its
center at the origin of an x axis. The rocket consists of a central block C (of mass M = 6kg
) and
blocks L and R (each of mass m= 2kg)
on the left and right sides. Small explosions can shoot
Fundamentals of Physics
Halliday & Resnic
東海大學物理系
either of the side blocks away from block C and along the x axis. Here is the sequence: (1) At time
t = 0 , block L is shot to the left with a speed of 3 m/s relative to the velocity that the explosion
gives the rest of the rocket. (2) Next, at time t = 0.8s,
block R is shot to the right with a speed of 3
m/s relative to the velocity that block C then has. At t = 2.8s,
what are (a) the velocity of block C
and (b) the position of its center?
圖 9-58 顯示了一個兩端的火箭,一開始被放在無摩擦平面上,中心在 x 軸原點。火箭包含
有中間的積木 C(質量 M = 6kg
),和左右兩個積木 L 和 R(質量都是 m= 2kg)。一個小爆
炸可以把兩個小積木沿著 x 軸推離積木 C。接著(1)在 t = 0 ,積木 L 以相對速度 3m/s 被
推到左邊,爆炸使得火箭靜止(2)接下來, t = 0.8s時,積木
R 以相對速度 3m/s 被推到積
木 C 的左邊,在 t = 2.8s,(a)問積木
C 的速度,和(b)中心位置?
(圖 9-58)
:(a) With SI units understood, the velocity of block L (in the frame of reference indicated in
the figure that goes with the problem) is (v1 – 3)i ^ . Thus, momentum conservation
(for the explosion at t = 0) gives
mL (v1 – 3) + (mC + mR)v1 = 0
3 mL 3(2 kg)
which leads to v1 =
mL + mC +
=
mR 10 kg
= 0.60 m/s.
Next, at t = 0.80 s, momentum conservation (for the second explosion) gives
mC v2 + mR (v2 + 3) = (mC + mR)v1 = (8 kg)(0.60 m/s) = 4.8 kg·m/s.
This yields v2 = – 0.15. Thus, the velocity of block C after the second explosion
is v2 = –(0.15 m/s)i ^ .
(b) Between t = 0 and t = 0.80 s, the block moves v1Δt = (0.60 m/s)(0.80 s) = 0.48 m.
Between t = 0.80 s and t = 2.80 s, it moves an additional
v2Δt = (– 0.15 m/s)(2.00 s) = – 0.30 m.
Its net displacement since t = 0 is therefore 0.48 m – 0.30 m = 0.18 m.
【Problem 9-45】
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the
explosion, one piece, of mass m, moves with velocity ( −30
m/ s) i and a second piece, also of mass
m, moves with velocity ( − 30 m/ s) j.
The third piece has mass 3m. Just after the explosion, what
are the (a) magnitude and (b) direction of the velocity of the third piece?
一個導管靜止在 xy 座標的原點,爆炸分成三塊。爆炸後,其中一塊質量 m,移動速度
( −30 m/ s) i。第二塊質量一樣是
m,速度 ( − 30 m/ s) j。第三塊質量
3m,問第三塊速度(a)
大小(b)方向?
:已知 m1m
v m s
i
= , 1 = ( −30
/ )