Question

A blue car with mass mc = 539 kg is moving east with a speed of vc = 15 m/s and collides with a purple truck with mass mt = 1300 kg that is moving south with an unknown speed. The two collide and lock together after the collision moving at an angle of ? = 52

Answer #1

**m _{c} = 539 kg**

**m _{t} = 1300 kg**

**V _{cxi} = 15 m/s** = initial velocity of
car in X-direction (east) going east.

**V _{txi}**= initial velocity of truck in
X-direction (east) = 0

**V _{cyi}** = initial velocity of car in
Y-direction (south) = 0

**V _{tyi}** = initial velocity of truck in
Y-direction (south)?

**V = final velocity**

**V _{fx}** = final velocity along
X-direction =

**V _{fy}**= final velocity along
Y-direction =

1) initial momentum of car = **m _{c}
V_{cxi}** = 539 x 15 =

2) using **conservation of momentum along the
x-direction** ::

**m _{c} V_{cxi} + m_{t}
V_{txi} = (m_{c} + m_{t})
V_{fx}**

539 x 15 + 1300 x 0 = (539 + 1300) V cos52

8085 = 1132.2 V

**V = 7.14 m/s**

So **V _{fy}**= V sin52 = 7.14 Sin52 =

using conservation of momentum along the y-direction ::

**m _{c} V_{cyi} + m_{t}
V_{tyi} = (m_{c} + m_{t})
V_{fy}**

539 x 0 + 1300 x V_{tyi} = (539 + 1300) **(5.63
m/s)**

**V _{tyi} = 7.96 m/s**

initial momentum of truck = **m _{t}
V_{tyi}**= 1300 x 7.96 =

3) speed of truck before collision = **V _{tyi} =
7.96 m/s**

4) final momentum of combination = **(m _{c} +
m_{t}) V = (1300 + 539) 7.14 =** 13130.5 kgm/s

5) V = 7.14 m/s

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