Question

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 afterward? PLEASE HELP

Answer 1

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 2.88 m/s.