質量0.2kg的球 - Ans.fyi 參考答案

質量0.2kg的球

2013-01-10 9:01 am

質量0.2kg的球以20m/s的水平速度撞到一面牆上距地面高20m處水平反跳後落於具牆基30m處的地面 (g=10m/s平方) 則

1.若球與牆接觸時間0.1杪則接觸期間球所受的平均作用力為若干

2.球落地前瞬間的動量大小若干

3.什麼是牆基

更新1:

球反彈後水平等速度運動 30=v*2 =15m/s 鉛直自由落體 s=vot+at平方/2 20=10/2*t平方 t=2 所以vx=15m/s vy=10*2=20m/s 1.F=ma =0.2*20-(-15)/0.1 =70牛噸 2.P=m*v =0.2*根號20平方+15平方 =5kg*m/s

回答 (1)

天同

2013-01-10 4:11 pm

✔ 最佳答案

1. Consider the vertical motion of the ball after rebound from the wall.
Apply equation: s = ut + (1/2)at^2
with s = -20 m, u = 0 m/s, a = -g(= -10 m/s^2), t =?
hence, -20 = (1/2).(-10).t^2
t = 2 s

Horizontal rebound velocity of ball = 30/2 m/s = 15 m/s

Change of of (horizontal) momentum during rebound
= 0.2 x [(-15) - 20] kg.m/s = -7 kg.m/s
Impact force = -7/0.1 N = -70 N

2. Vertical velocity of ball before striking the ground
= (-10) x 2 m/s = - 20 m/s
Hence, resultant velocity before striking the ground
= square-root[15^2 + (-20)^2] m/s = 25 m/s
Momentum = 0.2 x 25 kg.m/s = 5 m/s

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