A ball with mass m and velocity v hits a wall, and suffers an elastic colision. So it comes back vith velocity -v (1 dimension motion). Since linear momentum must be conserved, it means the wall has a momentum p=2mv after the ball comes back. How can we prove it? (mv=-mv+2mv)
\[p_1+p_2=p_3+p_4\] \[m_1v_1+m_2(0)=m_1v_{1f}+m_2v_{2f}\] \[m_1v_1=m_1v_{1f}+m_2v_{2f}\] \[v_{1f}=-v_1\] \[m_1v_1=-m_1v_1+m_2v_{2f}\] \[m_1v_1+m_1v_1=m_2v_{2f}\] \[2m_1v_1=m_2v_{2f}\]
I just find this problem a little silly because that wall is going to instantly stop after it's hit (unless this is a gigantic ball.) i guess you can see this as a wrecking ball hitting a wall.
more obvious if you think of it as a gigantic ball hitting a smaller wall
Was it? hahaha I guess that is fine, since linear momentum must be conserved. But it was a question by professor Walter Lewin, in 8.01 (MIT Physics), that he wanted his students to think about. I gonna watch it again and see if that was what he was asking for... At the moment it seemed as a brain teaser. Anyway, thanks haha. The question is really silly...
So, i guess the question is 'how can the wall have momentum and yet no kinect energy (since m=infinity and thus v=0)?" - an ideal case What is the meaning of this?
I haven't seen the video, but I think the important fact here is that whereas momentum varies in v, kinetic energy varies in v². If mass of wall M is much greater than mass of ball m, then MV will be finite, whereas MV²/2 will be infinitesimal.
momentum is conserved if no external force is applied on the system.in this case i believe an external force is present
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