Question

50 kg boy jumps off the front of a 1.5 kg skateboard moving forward. Find the skateboard's velocity

immediately after the boy jumps, assuming that the skateboard's initial velocity is 3.5 m/s and the boy's
velocity when jumping off the front is 5 m/s.

Answer 1

170.17m/s.

Step-by-step explanation:

To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of an isolated system remains constant if no external forces act on it. Mathematically, this can be expressed as:

m1u1+m2u2=m1v1+m2v2m1​u1​+m2​u2​=m1​v1​+m2​v2​

where:

m1m1​ and m2m2​ are the masses of the two objects,

u1u1​ and u2u2​ are their initial velocities,

v1v1​ and v2v2​ are their final velocities.

In this case, the boy and the skateboard form an isolated system, so the total momentum before the jump must be equal to the total momentum after the jump.

Let:

m1m1​ be the mass of the boy (50 kg),

u1u1​ be the initial velocity of the boy (5 m/s),

m2m2​ be the mass of the skateboard (1.5 kg),

u2u2​ be the initial velocity of the skateboard (3.5 m/s),

v1v1​ be the final velocity of the boy (which is 0, as he jumps off),

v2v2​ be the final velocity of the skateboard (what we're trying to find).

The conservation of momentum equation becomes:

(50 kg×5 m/s)+(1.5 kg×3.5 m/s)=(50 kg×0)+(1.5 kg×v2)(50kg×5m/s)+(1.5kg×3.5m/s)=(50kg×0)+(1.5kg×v2​)

Solving for v2v2​:

250 kg⋅m/s+5.25 kg⋅m/s=0+1.5 kg⋅v2250kg⋅m/s+5.25kg⋅m/s=0+1.5kg⋅v2​

255.25 kg⋅m/s=1.5 kg⋅v2255.25kg⋅m/s=1.5kg⋅v2​

v2=255.25 kg⋅m/s1.5 kgv2​=1.5kg255.25kg⋅m/s​

v2≈170.17 m/sv2​≈170.17m/s

Therefore, the skateboard's velocity immediately after the boy jumps is approximately 170.17 m/s170.17m/s.