Question

A 2150 kg car, moving East at 10 m/s, collided and joins with a 3250 kg car. The cars move East together at 5.22 m/s. What is the 3250 kg car’s initial velocity calculated to the nearest tenth?

Answer 1

2.1 m/s

Step-by-step explanation:

Momentum is conserved hence sum of initial momentum equals the sum of final momentum

`M_1* v_1 + M_2* v_2= (M_1+ M_2)* V_c`

where
`M_1` is the mass of 2150 Kg car,
`M_2` is the mass of 3250 Kg car,
`V_1` is the velocity of the 2150 Kg car,
`V_2` is the velocity of the 3250 Kg car,
`V_c` is the common velocity

Making
`V_2` the subject then

`v_2=\frac {(M_1+ M_2)* V_c -M_1* v_1 }{M_2}`

Substituting 2150 Kg for
`M_1`, 3250 Kg for
`M_2`, 10 m/s for
`v_1`, 5.22 m/s for
`v_c` then we obtain

`v_2=\frac {(2150 kg+3250 kg)* 5.22 m/s- 2150 kg* 10 m/s}{3250}=2.057846154 m/s\approx 2.1 m/s`