two blocks are initially at rest with a compressed spring between them. when the spring is released, the 2m…
two blocks are initially at rest with a compressed spring between them. when the spring is released, the 2m block moves to the left and the m block moves to the right. this represents an inelastic collision in a closed system. what best describes the blocks after the spring is released? the total energy is zero. the total momentum is zero. the momentum of the m block is zero. the momentum of the 2m block is zero.
Answer
Explanation:
Step1: Recall conservation of momentum
In a closed - system, the initial momentum is zero since the blocks are initially at rest ($p_i = 0$). According to the law of conservation of momentum $p_i=p_f$, where $p_i$ is the initial momentum and $p_f$ is the final momentum.
Step2: Analyze final momentum
The two - block system is a closed system. After the spring is released, the $2m$ block moves to the left with momentum $p_{2m}=- 2mv_{2m}$ (negative because it moves to the left) and the $m$ block moves to the right with momentum $p_{m}=mv_{m}$. Since $p_i = 0$, we have $p_f=p_{2m}+p_{m}=0$, or $-2mv_{2m}+mv_{m}=0$, which implies $mv_{m}=2mv_{2m}$. The total energy is non - zero (it is the initial elastic potential energy of the compressed spring which is converted into kinetic energy of the two blocks). The momentum of each individual block is non - zero.
Answer:
The total momentum is zero.