# Video: Determining the Net Motion of an Object Accelerated Perpendicular to Its Velocity Direction

A hockey puck of mass 150 g is sliding due east on a frictionless table with a speed of 10 m/s. Suddenly, a constant force of magnitude 5.0 N and direction due north is applied to the puck for 1.5 s. Find the northward component of the hockey puck’s momentum at the end of the 1.5-s interval. Find the eastward component of the hockey puck’s momentum at the end of the 1.5-s interval.

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### Video Transcript

A hockey puck of mass 150 grams is sliding due east on a frictionless table with a speed of 10 meters per second. Suddenly, a constant force of magnitude 5.0 newtons and direction due north is applied to the puck for 1.5 seconds. Find the northward component of the hockey puck’s momentum at the end of the 1.5-second interval. Find the eastward component of the hockey puck’s momentum at the end of the 1.5-second interval.

We can refer to the northward component of the puck’s momentum and the eastward component of the puck’s momentum at the end of this interval as 𝑝 sub 𝑛 and 𝑝 sub 𝑒, respectively. So, we want to solve for 𝑝 sub 𝑛 and 𝑝 sub 𝑒, the northward and eastward components of the puck’s momentum , after a time, we’ve called Δ𝑡, of 1.5 seconds has elapsed.

Before this time interval begins, we’re told that the puck is moving to the east with a speed of 10 meters per second. At time 𝑡 equals zero, a force 𝐹 is applied to the puck and lasts 1.5 seconds. Regarding momentum, there are two relationships we can recall. The first is that an object’s momentum is equal to its mass times its velocity. And the second is that the change in an object’s momentum is equal to the net force acting on it times the time over which that force acts.

Considering the northward component of our puck’s momentum, we know that at the outset, it has no momentum in that direction because its velocity to the north is zero. But after 1.5 seconds of the force 𝐹 being applied pointing to the north, referring to our second relationship for momentum, we can write that 𝑝 sub 𝑛 is equal to the magnitude of 𝐹 multiplied by Δ𝑡.

This change in momentum to the north is equal to the final northward momentum component because the original component was zero. Plugging in for these two values, we find that 𝑝 sub 𝑛 is equal to 7.5 kilograms meters per second. That’s the northward component of the puck’s momentum after a time Δ𝑡 has passed.

Next, we wanna solve for the eastward component of the momentum after that same time interval. Unlike to the north, to the east our puck does have an initial speed. And because the force applied to the puck is perpendicular to that direction, it doesn’t affect the puck’s speed to the east. This means the puck’s momentum to the east is a constant value. It doesn’t change in time and is always equal to the puck’s mass times 𝑣 sub 𝑖.

We’re told that the mass of the puck is 150 grams. But when we enter it into our expression, we write it in units of kilograms to be consistent with the units in the rest of the expression. Calculating this product, we find that 𝑝 sub 𝑒 is 1.5 kilograms meters per second. That’s the puck’s eastward momentum component after Δ𝑡 has passed or at any time.