1.一輛質量為4000kg的貨車以17m/s的速率向北行駛,並跟一輛私家車迎頭相撞。私家車的質量為1500kg,相撞前正以20m/s的速率行駛。相撞後,私家車以10m/s的速率向北行駛。
a.求貨車相撞後的速度。5.75m/s(北)
b.如果在相撞時作用於私家車的力為300000N,碰撞時間是多少?0.15s
2.木塊置於平滑的桌面上。竟成向木塊發射一顆質量為50g的子彈,子彈的初速度為
150m/s。子彈擊中並嵌入木塊內。已知木塊的質量為2kg。
a.求子彈的初動量。7.5kgms(右)
b.求作用於木塊的衝量。7.32kgms
3.皮球A和B在一塊平滑的表面上移動,皮球A向右移,皮球B向左移,它們的速率分别是
5m/s和2m/s,質量分别是50g和30g。球相撞後,B向相反方向移動,但速率仍是2m/s。
a.求碰撞後A的速度。2.6m/s(右)
b.求在碰撞過程中失去的動能。0.456J
4.在光光滑的桌面上有三個球A、B和C,它們的質量相同。開始時球都是靜止的。小明先擊打A,使它與B相撞,B再撞向C。A和B碰撞後,兩者的速度是3m/s,且與A原本的移動方向成20度。
a.求A的初速率。5.64m/s
b.假設B撞向C後,便立即停下來。求C在碰撞後的速率。3m/s
急!!!!求教幾條物理題目 20點!!!!!
2015-05-04 1:43 am
回答 (2)
2015-05-04 2:57 am
✔ 最佳答案
1(a) Use conservation of momentum (take north direction as +ve)4000 x 17 + 1500 x (-20) = 4000v + 1500 x 10
where v is the speed of the lorry after colision
solve for v gives v = 5.75 m/s towards north.
(b) Change of momentum of the car = 1500 x (10 - (-20)) kg.m/s = 45 000 kg.m/s
Hence, impact time = 45000/300000 s = 0.15 s
2(a) Initial momentum = (50/1000) x 150 kg.m/s = 7.5 kg.m/s
(b) Use conservation of momentum
0.05 x 150 = (0.05 + 2)v
where v is the speed of the wooden block after collision
i.e. v = 3.659 m/s
Change of momentum of wooden block = 2 x 3.659 kg.m/s = 7.317 kg.m/s
Hence, impulse acting on the block = 7.317 N.s
3(a) Use conservation of momentum (take direction to the right as +ve)
0.05 x 5 + 0.03 x (-2) = 0.05v + 0.03 x (2)
where v is the velocity of ball A after collision
v = 2.6 m/s towards the right
(b) Kinetic energy before collision
= (1/2).(0.05).(5^2) + (1/2).(0.03).(2^2) J = 0.685 J
Kinetic energy after collision
= (1/2).(0.05).(2.6^2) + (1/2).(0.03).(2^2) J = 0.229 J
Hence, loss of kinetic energy = (0.685 - 0.229) J
= 0.456 J
2015-05-03 19:16:50 補充：
4(a) Let u be the initial velocity of ball A, and m be the mass of each ball.
Use conservation of momentum
mu = m(3.cos(20)) + m(3.cos(20))
i.e. u = 2 x 3.cos(20) m/s = 5.64 m/s
2015-05-03 19:18:49 補充：
(b) Use conservation of momentum on balls B and C
m.(3) + m(0) = m(0) + mv
where v is the velocity of ball C after collision
hence, 3m = mv
i.e. v = 3 m/s
2015-05-04 6:29 am
1a
m1u1+m2u2=m1v1+m2v2
4000(17)+1500(-20)=4000v2+1500(10)
v1=5.75 m/s
1b
作用於私家車的力=作用於貨車的力=300000
F=ma
300000=4000[(17-5.75)]/t
t=0.15(s)
2a
初動量=(50/1000)(150)
初動量=7.5kg m/s
2b
m1u1+m2u2=m1v1+m2v2
(50/1000)(150)+2(0)=[(50/1000)+2]v
v=3.658536585
F=ma
F=2*3.658536585
F=7.32 N
3a
m1u1+m2u2=m1v1+m2u2
(50/1000)(5)+(30/1000)(-2)=(50/1000)v+(30/1000)(2)
v=2.6 m/s
3b
E=0.5(50/1000)(5)^2 +0.5(30/1000)(2)^2 -0.5(50/1000)(2.6)^2 -0.5(30/1000)(2)^2
E=0.456J
4a
m1u1+m2u2=m1v1+m2v2
u=(3+3)cos20
u=5.64 m/s
4b
m1u1+m2u2=m1v1+m2v2
3+0=0+v
v=3 m/s
m1u1+m2u2=m1v1+m2v2
4000(17)+1500(-20)=4000v2+1500(10)
v1=5.75 m/s
1b
作用於私家車的力=作用於貨車的力=300000
F=ma
300000=4000[(17-5.75)]/t
t=0.15(s)
2a
初動量=(50/1000)(150)
初動量=7.5kg m/s
2b
m1u1+m2u2=m1v1+m2v2
(50/1000)(150)+2(0)=[(50/1000)+2]v
v=3.658536585
F=ma
F=2*3.658536585
F=7.32 N
3a
m1u1+m2u2=m1v1+m2u2
(50/1000)(5)+(30/1000)(-2)=(50/1000)v+(30/1000)(2)
v=2.6 m/s
3b
E=0.5(50/1000)(5)^2 +0.5(30/1000)(2)^2 -0.5(50/1000)(2.6)^2 -0.5(30/1000)(2)^2
E=0.456J
4a
m1u1+m2u2=m1v1+m2v2
u=(3+3)cos20
u=5.64 m/s
4b
m1u1+m2u2=m1v1+m2v2
3+0=0+v
v=3 m/s
收錄日期: 2021-04-15 19:14:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150503000051KK00064