Question

in a very busy off - campus eatery one chef sends a 211 g pizza sliding down the counter from left to right at 1.69 m/s. almost simultaneously, the other chef launches a 345 g veggieburger along the same counter from right to left at 2.37 m/s. the two delicacies collide head - on at the given speeds and merge together.
the counter is practically friction - free due to accumulated grease.
at what speed ( v ) does it move?
in what direction does the merged meal move, if it moves at all?
does not moveto the rightto the left

Explanation:

Step1: Determine the initial momenta

Let the right - hand direction be positive.
The mass of the pizza \(m_1 = 211g=0.211kg\), and its velocity \(v_1 = 1.69m/s\). So its initial momentum \(p_1=m_1v_1=(0.211)(1.69)\) \(kg\cdot m/s\approx0.3576kg\cdot m/s\)
The mass of the veggie - burger \(m_2 = 345g = 0.345kg\), and its velocity \(v_2=- 2.37m/s\). So its initial momentum \(p_2=m_2v_2=(0.345)(-2.37)\) \(kg\cdot m/s\approx - 0.8177kg\cdot m/s\)

Step2: Use the law of conservation of momentum

The total initial momentum \(P = p_1 + p_2\)
\(P=0.3576-0.8177=-0.4601kg\cdot m/s\)
After the collision, the combined mass \(M=m_1 + m_2=0.211 + 0.345=0.556kg\)
By the law of conservation of momentum \(P = Mv\), so \(v=\frac{P}{M}\)
\(v=\frac{- 0.4601}{0.556}\approx - 0.83m/s\)

Answer:

The merged meal moves to the left.
\(v = 0.83m/s\)