Question #96637
A shopper is in a supermarket pushing their unladen trolley westwards across an aisle at a
constant speed of 1.4 ms-1. Another shopper with groceries in their trolley is moving north out of
the aisle at 1.8 ms-1 and the trollies collide, the shoppers releasing the trollies in surprise at the
moment of collision. The trollies jam together and end up moving at a velocity of 1.234 ms-1 in a
direction 25.7° W of N. If 41.75 J of energy is lost into noise, deforming the trolleys etc. during
the collision, what is the mass of a trolley, (assuming they are the same type of trolley) and what
was the mass of the second shopper’s groceries?
1
Expert's answer
2019-10-25T11:31:38-0400

From the conservation of momentum:


m1v1=(m1+m2)vsin25.7°m_1v_1=(m_1+m_2)v\sin{25.7\degree}

m2v2=(m1+m2)vcos25.7°m_2v_2=(m_1+m_2)v\cos{25.7\degree}

m1m2=v2v1tan25.7°\frac{m_1}{m_2}=\frac{v_2}{v_1}tan{25.7\degree}

We have:


0.5m1v12+0.5m2v22=0.5(m1+m2)v2+E0.5m_1v_1^2+0.5m_2v_2^2=0.5(m_1+m_2)v^2+E

m1m2v12+v22=(1+m1m2)v2+2Em2\frac{m_1}{m_2}v_1^2+v_2^2=(1+\frac{m_1}{m_2})v^2+\frac{2E}{m_2}

1.81.4tan25.7°1.42+1.82=(1+1.81.4tan25.7°)1.2342+2(41.75)m2\frac{1.8}{1.4}tan{25.7\degree}1.4^2+1.8^2=(1+\frac{1.8}{1.4}tan{25.7\degree})1.234^2+\frac{2(41.75)}{m_2}

m2=42.0 kgm_2=42.0\ kg

The mass of a trolley:

m1=(42.0)1.81.4tan25.7°=26.0 kgm_1=(42.0)\frac{1.8}{1.4}tan{25.7\degree}=26.0\ kg

The mass of the second shopper’s groceries:


m2m1=42.026.0=16.0 kgm_2-m_1=42.0-26.0=16.0\ kg


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Comments

Assignment Expert
25.10.19, 18:32

Dear ArtisansCritic, thank you for your comment. Please see fixed solution

ArtisansCritic
25.10.19, 02:49

How do you get m1/m2=tan25.7?

Assignment Expert
23.10.19, 16:58

Dear Antikieran, you need to divide both sides of the equation on (0.5m2)

Antikieran
23.10.19, 12:33

How do you explain the step between 0.5m1v1^2.... and m1/m2?

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