郑州市2023-2024学年上学期期末考试高中一年级物理参考答案一．选择题（共12小题，共48分）1．C2．B3．A4．B5．D6．D7．D8．C9．BD10．AC11．BC12．AD二．实验题（共2小题，共16分）13．（1）CD（2）F′；（3）2。14．（1）木板右端垫起的高度较小（即平衡摩擦力不足或补偿阻力不足）；砂桶和沙子的总质量m不满足远小于小车质量M(或砂桶和沙子的重力与绳子的拉力差别较大)；（）（2）1.6；3.0（3）????−�+��三．计算题（共4小题，共36分�）�15.（8分）解：（1）雪橇匀速下滑时，由平衡条件可得：沿斜面方向：=mgsin37°························1′垂直斜面方向：=mg························1′�f又因为=········0············1′（注：mgsin37°=mg·······3′）��cos37联立解得：＝0.75························1′0�fμ��μcos37雪橇沿水平面做匀速直线运动，根据平衡条件可得：μ水平方向：································1′竖直方向：0mg································1′�cos37=�f0又因为=��+�···s··i·n··3··7·····=··········1′（注：=（mg-）··········3′）00联立解得�f：μ�=�360N································1�′cos37μ�sin37．（分）解：（）对乙车，匀速运动的位移′168�1···························1由得：匀减速运动的位移12′2�··=····�····∆··�···············12�22乙车�的=总2�位�移x=+=50m····�····=····2··�·············1′12（注：写乙车的位�移�′）x=+2··········2�22（2）设乙车减速t1时间�恰∆能�追2�上甲车，则有：-=a'···························2′�2�1�（1）（）···························2′1212111联立解�得∆：�a+'＝22.5�m/+s2�·�····=····�··+····�·······∆··�··+·1′�（注：x甲=（）·····1′x乙（）·····1′）117.（9分）解�：1（1∆）�+物�块1在斜面上下滑时=，�2∆�+2�1+�2�1由牛顿第二定律得：mgsin370-µmgcos370=ma·················1′由运动学公式得：···························1′联立解得：24m/s······2·····················1′�=2���1=2��1（2）令物块在斜面上运动的时间为t1，由=at1得：t1=2s···························1′�1=�1第1页（共2页）物块运动到传送带上时，由牛顿第二定律得：2mg＝ma2················1′物块与传送带共速时，由速度公式得：＝﹣′vv1μa2t2···························1解得：t2＝0.5s物块匀减速运动的位移t2（或=v1t2-）解得：x=1.5m······················1′�1+�1222匀速运动的时间t3=�=解2得：t3=�1s2··�···�······················1′�2−�物块运动的总时间t�t1+t2+t3=3.5s···························1′18.（9分）解：（1）力F作用时，假设M和m发生相对运动，=2对m有：µ1mg=mamam=3m/s···························1′2对M有：F-µ2（M+m）g-µ1mg=MaMaM=9m/s···························1′2由此可知，假设成立，长木板的加速度aM为9m/s（2）力F作用时间内，对m有：xm=解得：xm=1.5m···························1′12对M有：=解得：2=�4�.5�1m···························1′12物块到长木M板1右端�距离M1′�2��1-x�m=3m···························1（3）在t1时间内，由v=at得，vm=3m/svM=9m/s∆�=�M1撤去F后，M做匀减速直线运动，m做匀加速直线运动2对m有：µ1mg==3m/s对有：（'）'2′Mµ2M�+m�mg+µ1m�mg=M=3m/s···························1经时间两者共速，则有''解得′tvm+t�=M-t�Mt=1s···························1'2'对M，=t-t�=m7.5m�M�M···························1′1'当两者共M速2后M，两者相M对静止M，2对和整体有：（）（）′��2��MmM+ma共=µ2M+mg···················12解得：a共=1m/s2对整体，由v=2a共x解得：=18m···························1′长木板的总位移=++�M3解得：=30m···························1′�M�M1�M2�M3�M第2页（共2页）