Physics Help Forum Collision problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 May 13th 2017, 12:10 PM #1 Junior Member   Join Date: May 2017 Posts: 2 Collision problem Hey, guys. I have problems with this exercise. https://ibb.co/b4PiQk I thought that the linear momentum would remain conserved and then i wrote the equation for the conservation of the total mechanical energy. But I don't know how to continue because actually I didnt understand what the problem asked me to find. Can you help me please?
 May 13th 2017, 12:38 PM #2 Senior Member   Join Date: Jun 2010 Location: NC Posts: 363 Hey Aledg... The image you posted is rotated 90deg CCW. Text aligned vertical. Hard to read. Here's all I have close to what you seek. Box Bounces Between Springs | THERMO Spoken Here! Good luck, JP
 May 13th 2017, 01:25 PM #3 Junior Member   Join Date: May 2017 Posts: 2 Ok sorry Here it is https://ibb.co/nvbGBQ
 May 13th 2017, 02:00 PM #4 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 688 I don't look at unknown sites off forum, especially since the recent cyber attack. However Thermo's reply seems to indicate that you have springs involved. Why do you think momentum is conserved in this case? Momentum is not conserved in elastic collisions. https://www.google.co.uk/search?hl=e...73.VWvRINT2kBM
May 13th 2017, 04:58 PM   #5
Senior Member

Join Date: Nov 2013
Location: New Zealand
Posts: 521
 Originally Posted by studiot I don't look at unknown sites off forum, especially since the recent cyber attack.
Which image hosting sites does physhelpforum say are safe?

 May 14th 2017, 08:46 AM #6 Senior Member     Join Date: Aug 2008 Posts: 110 when the block achieves maximum compression of the spring, both the block and platform are at the same velocity, $v_1$ ... momentum is conserved ... $mv_0 = (m+M)v_1 \implies v_1 = \dfrac{mv_0}{m+M} = \dfrac{v_0}{41}$ all forces are conservative, so energy is conserved ... $\dfrac{1}{2}mv_0^2 = \dfrac{1}{2}(m+M)v_1^2 + \dfrac{1}{2}k(x_{max})^2$ substituting given values for $m$, $M$, and $k$ ... $x_{max} = \dfrac{v_0}{5\sqrt{82}}$ after the instant of maximum spring compression, momentum is conserved ... $(m+M)\cdot \dfrac{v_0}{41} = mv_2 + Mv_3$ where $v_2$ is the cube's final velocity and $v_3$ is the platform's final velocity ... note both velocities are relative to an outside observer. since energy is conserved, opening speed after the collision = closing speed before the collision ... $v_3-v_2 = v_0 - 0$ $v_3 = v_0+v_2$ $(m+M)\cdot \dfrac{v_0}{41} = mv_2 + M(v_0+v_2)$ $\dfrac{v_0}{2} = 20.5v_2 + 20v_0$ $v_2 = -\dfrac{39v_0}{41} \implies v_3 = \dfrac{2v_0}{41}$

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