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质量为3千克的球以3米/秒的速度滚动并与质量为1千克的静止球弹性碰撞。什么是球的碰撞后速度?
能量和动量守恒方程。 u_1'= 1.5m / s u_2'= 4.5m / s正如维基百科建议的那样:u_1'=(m_1-m_2)/(m_1 + m_2)* u_1 +(2m_2)/(m_1 + m_2)* u_2 = =(3- 1)/(3 + 1)* 3 +(2 * 1)/(3 + 1)* 0 = = 2/4 * 3 = 1.5m / s u_2'=(m_2-m_1)/(m_1 + m_2) * u_2 +(2m_1)/(m_1 + m_2)* u_1 = =(1-3)/(3 + 1)* 0 +(2 * 3)/(3 + 1)* 3 = = -2 / 4 * 0 + 6/4 * 3 = 4.5m / s [方程式来源]推导动量和能量状态守恒:动量P_1 + P_2 = P_1'+ P_2'因为动量等于P = m * u m_1 * u_1 + m_2 * u_2 = m_1 * u_1'+ m_2 * u_2' - - - (1)能量E_1 + E_2 = E_1'+ E_2'因为动能等于E = 1/2 * m * u ^ 2 1/2 * m_1 * u_1 ^ 2 + 1/2 * m_2 * u_2 ^ 2 = 1/2 * m_1 * u_1 ^ 2'+ 1/2 * m_2 * u_2 ^ 2' - - - (2)您可以使用(1)和(
质量为2千克的球以9米/秒的速度滚动并与质量为1千克的静止球弹性碰撞。什么是球的碰撞后速度?
无取消(v_1 = 3 m / s)无取消(v_2 = 12 m / s)两个物体碰撞后的速度见下文解释:颜色(红色)(v'_1 = 2.64 m / s,v' _2 = 12.72 m / s)“使用动量对话”2 * 9 + 0 = 2 * v_1 + 1 * v_2 18 = 2 * v_1 + v_2 9 + v_1 = 0 + v_2 v_2 = 9 + v_1 18 = 2 * v_1 + 9 + v_1 18-9 = 3 * v_1 9 = 3 * v_1 v_1 = 3 m / s v_2 = 9 + 3 v_2 = 12 m / s因为有两个未知我不知道你怎么能解决上述问题不使用,动量守恒和能量守恒(弹性碰撞)。两者的组合产生2个等式和2个未知,然后你解决:“动量”的守恒:m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2 =======>(1)设,m_1 = 2kg; m_2 = 1千克; V_1 =9米/秒; v_2 = 0m / s能量守恒(弹性碰撞):1 / 2m_1v_1 ^ 2 + 1 / 2m_2v_2 ^ 2 = 1 / 2m_1v'_1 ^ 2 + 1 / 2m_2v'_2 ^ 2 =======>(2我们有2个方程和2个未知数:从(1)==> 2 * 9 = 2v'_1 + v'_2;颜色(蓝色)(v