Help Needed: Calculating Mutual Speed After Tackling Collision

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To calculate the mutual speed after a head-on collision between a 139 kg tackler moving at 2.53 m/s and an 87.4 kg halfback moving at 5.14 m/s, the inelastic collision formula Vf = (m1v1 + m2v2)/(m1 + m2) is used. The velocities must account for direction, with one set as negative. After performing the calculations, the mutual speed immediately after the collision is determined to be 3.54 m/s. This result reflects the combined momentum of both players post-collision. Understanding these principles is essential for accurate physics problem-solving.
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could someone help me with this problem?

A 139 kg tackler moving at 2.53 m/s meets head-on (and tackles) a 87.4 kg halfback moving at 5.14 m/s. What will be their mutual speed immediately after the collision?

Oh right. Here's what I've tried doing but no luck.

m1 = 139kg V1 = 2.53 m/s
m2 = 87.4 kg V2 = 5.14 m/s

m1v1+msv2 = Vf(mi+m2)

Vf = (m1v1+m2v2)/(m1+m2)
 
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Nevermind I got it. Seems I needed to set one of the velocities as negative since it was an inelastic collision.
 


Hi there, it looks like you're on the right track with your calculations. To find the mutual speed after the collision, you can use the formula you mentioned: Vf = (m1v1 + m2v2)/(m1 + m2). This formula takes into account the masses and velocities of both objects before the collision to determine their combined speed after the collision.

Let's plug in the values given in the problem:
Vf = (139 kg * 2.53 m/s + 87.4 kg * 5.14 m/s)/(139 kg + 87.4 kg)
Vf = (351.67 kg*m/s + 449.236 kg*m/s)/226.4 kg
Vf = 800.906 kg*m/s/226.4 kg
Vf = 3.54 m/s

So the mutual speed after the collision is 3.54 m/s. I hope this helps! Keep up the good work in your calculations.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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