[Grade 12 physics momentum] How can I determine the speed at which the car entered the intersection?

Question

The question reads:
'In a certain road accident a car of mass 2.0*10^3 kg, travelling south, collided in the middle of an intersection with a truck of mass 6.0*10^3 kg, travelling west. The vehicles locked together and slid freely (with neglibible friction) off the road along a line point south west. The truck entered the intersection going 8.9 m/s.
A) determine the speed at which the car entered the intersection'

I drew a vector diagram and wrote down all the information, but I feel like I don't have enough. Could somebody guide me through this?

Steve4Physics · Accepted Answer

6.0x10³ x 8.9 =  2.0x10³V
V = 6.0x10³ x 8.9/(2.0x10³)
. . = 26.7m/s
_________________________

If you want a full explanation:

Car's initial velocity = V (in south direction)
Car's initial momentum = 2.0x10³V south

Truck's initial momentum = 6.0x10³ x 8.9

For brevity, call the locked-together vehicles 'L'.
From conservation of momentum, L’s south component of momentum =  2.0x10³V
From conservation of momentum, L’s west component of momentum =    6.0x10³ x 8.9

'South-west' means 45° from south and 45° from west. So, because of the symmetry, L’s momentum has equal south and west components. 
2.0x10³V =  6.0x10³ x 8.9
V =  6.0x10³ x 8.9/2.0x10³
. . = 26.7m/s

Some Body · Answer

..m, v
....↓
....╬ ← M, V
↙
m+M, u

Conservation of momentum in x direction:
MV = (m+M) u cos 45°

Conservation of momentum in y direction:
mv = (m+M) u sin 45°

To solve for v, isolate u in the first equation and substitute into the second:
u = MV / [ (m+M) cos 45° ]

mv = (m+M) MV / [ (m+M) cos 45° ] sin 45°
mv = MV tan 45°
v = (M/m) V

Given:
m = 2.0×10^3 kg
M = 6.0×10^3 kg
V = 8.9 m/s

Plug in and solve.

Trending News

[Grade 12 physics momentum] How can I determine the speed at which the car entered the intersection?

2 Answers