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rjkz [21]
3 years ago
15

A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

erward? PLEASE HELP
Physics
1 answer:
Ierofanga [76]3 years ago
7 0

2.88 m/s is the velocity afterward.

Explanation:

By using the law of conservation of momentum

Initial momentum = final momentum

\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \times \mathrm{v} \text { equation }(1)

\mathrm{m}_{1}=91.5 \mathrm{kg} \text { is the mass of the first player }\mathrm{m}_{2}=63.5 \mathrm{kg}

\mathrm{m}_{2}=63.5 \mathrm{kg} \text { is the mass of the second player }

\mathrm{u}_{1}=2.73 \mathrm{m} / \mathrm{s} \text { is the initial velocity of the first player (choosing east as positive direction) }

\mathrm{u}_{2}=3.09 \mathrm{m} / \mathrm{s} \text { is the initial velocity of the second player }

v = is their combined velocity afterwards

Solving equation (1) for v

V=\frac{m_{1} u_{2}+m_{2} u_{2}}{m_{1}+m_{2}}

\mathrm{V}=\frac{(91.5 \times 2.73)+(63.5 \times 3.09)}{91.5+63.5}

\mathrm{V}=\frac{(249.7+196.2)}{155}

\mathrm{V}=\frac{445.9}{155}

V = 2.88 m/s

Therefore the velocity afterward is <u>2.88 m/s</u>.

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