# A 50.0kg object is moving east at an unknown velocity when it collides with a 60.0kg stationary object. After the collision, the 50.0kg is traveling at a velocity of 6.0m/s 50.0degree Nof E and the...

A 50.0kg object is moving east at an unknown velocity when it collides with a 60.0kg stationary object. After the collision, the 50.0kg is traveling at a velocity of 6.0m/s 50.0degree Nof E and the 60.0kg object is traveling at a velocity of 6.3m/s 38 degree Sof E. What was the velocity of the 50.0kg object before the collision?

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This problem can be solved by using the definition of the linear momentum: `vecp = m*vecV` (mass times velocity) and the law that momentum of a system in the absense of external forces is conserved.

According to the conservation of linear momentum law, the vector linear momentum of the 50 kg object before the collision is the same as the total vector linear momentum of 50 kg and 60 kg objects after the collision. Consider the components of the vector momentum along the east-west line before and after the collision:

Before the collision the 60 kg object is stationary, it has no linear momentum. The total momentum is

`p_1 = 50kg*V` , where V is the velocity of 50 kg object before the collision.

After the collision the component of total momentum of both objects along east-west line is

`p_2 = 50kg*6 m/s *cos50 + 60kg*6.3 m/s*cos38 = 490.7 kg*m/s`

Since `p_1 = p_2` , `50kg*V = 490.7 kg*m/s` and

`V=9.8 m/s`

**The velocity of 50 kg object before the collision is 9.8 m/s.**

**Sources:**

Before collission:

m=50kg, v=? , p=490.674 Kg m/s east (calculated below)

After collission:

For 50 kg object -

m=50kg, v=6m/s `50^o` N of E, p=300 Kg m/s `50^o` N of E

For 60 kg object-

m=60kg, v=6.3m/s `38^o` S of E, p=378 Kg m/s `38^o` S of E

Find horizontal and vertical components of 300 Kg m/s `50^o` N of E

`p_y=p sin theta`

`rArr p_y=(300Kg* m/s)(sin 50^o)`

`rArr p_y=229.8 Kg *m/s`

`p_x=pcos theta`

`rArr p_x=(300Kg* m/s)(cos 50^o)`

`rArr p_x=192.8` Kg m/s

Find horizontal and vertical components of 378 Kg m/s `38^o` S of E

`38^o S of E=378^o`

`p_y=p sin theta`

`rArr p_y=(378Kg* m/s)(sin 322^o)`

`rArr p_y=-232.697` Kg m/s

`p_x=pcos theta`

`rArr p_x=(378Kg* m/s)(cos 322^o)`

`rArr p_x=297.864` Kg m/s

`Sigma p_y=-2.8968` Kg m/s

`Sigma p_x=490.674` Kg m/s

Now add `Sigma p_y` and `Sigma p_x` using `p=sqrt((p_x)^2+(p_y)^2)`

So,` p=sqrt((490.674 Kg* m/s)^2+(2.8968 Kg* m/s)^2)`

`=490.6826` Kg m/s

`p=m*v`

`rArr v=p/m` =`(490.6826 Kg *m/s)/(50kg)`

`rArr v=9.8 ` m/s east.

**Therefore, the velocity of 50 kg object before the collision is 9.8 m/s east.**

**Sources:**