(1)在0-1.0s时间内小球沿x轴正方向做匀加速运动,则有 a 1=
E 1 q
m =
2×1 0 6 ×1×1 0 -10
1×1 0 -3 m/s 2=0.2m/s 2,v 1=a 1•△t 1=0.2×1.0m/s=0.2m/s;在0-1.0s时间内小球的位移为x 1=
1
2 a 1 (△t ) 2 =0.1m,小球从原点运动到(0.1m,0)处;
(2)在t 1=1.0s到t 2=2.0s的时间内,x 2=x 1+
1
2 a 1△t 2 2+v 1△t 2=0.1m+
1
2 ×0.2×1.0 2m+0.2×1m=0.4m,y 2=
1
2 a 2△t 2 2=
1
2 ×0.2×1.0 2m=0.1m.
(3)由v 2 2=2v 1 2,得t 2=2.0s时小球的速度为:v 2=
2 v 1=
2 ×0.2m/s=0.28m/s,a 3=
v 2
△ t 3 =
0.28
1.0 m/s 2=0.28m/s 2,
E 3=
m a 3
q =
1×10-3×0.28
1×1 0 -10 =2.8×10 6V/m
(4)带电小球的运动情况如下:在0-1.0s时间内小球的位移为x 1=
1
2 a 1 (△t ) 2 =0.1m,小球从原点运动到(0.1m,0)处;在t 1=1.0s到t 2=2.0s的时间内,小球的轨迹是抛物线,从(0.1m,0)处运动到(0.4m,0.1m)处;在t 2=2.0s到t 3=3.0s的时间内,小球运动的位移为x 3=
1
2 a 3 (△t ) 2 =0.28m,从(0.4m,0.1m)处沿直线运动到(0.4m,0.2m)处;画出三段运动轨迹如图,
答:(1)在t 1=1.0s时小球的速度v 1的大小为0.2m/s;
(2)在t 2=2.0s时小球的位置坐标x 2为0.4m,y 2为0.1m.
(3)匀强电场E 3的大小为2.8×10 6V/m;
(4)绘出该小球在这3s内的运动轨迹如图所示.